Mathematical Text-Books. 



PRICE-LIST, JANUARY i, 1896. 



I. A DRlLL=BOOK IN TRIGONOMETRY. 

FIRST EDITION. 

For both the elementary and the advanced work of high-school and college 
clu.sscs. 

12010, cloth, xvi + 192 pp. Single copies by mail, $1.00. 

All cash orders (carriage at the buyer's cost), 75 cents. 



II. A DRILL=BOOK IN ALGEBRA. 

THIRD ED ITr ON. 



For the r 
lower clas 




ies and the 


A X 
A t 


LIBRARY OF CONGRESS, 


its. 


Eighteen 
in the clas 
fine stron 


: %j8A^|rig^lo 

Shelf ...Jlij 


ons, for use 
, clear type, 




UNITED STATES OF AMERICA. 





Two tables : one of the logarithms of three-figure numbers, the other of trigo- 
nometric ratios and their logarithms for angles differing by ten minutes. 

1 2mo, paper, 8 pp. Single copies by mail, 5 cents. 

All cash orders (carriage at the buyer's cost), 4 cents. 



V. FIVE=PLACE LOGARITHMS. 

IN PREPARATION. 



VI. PROBLEMS IN ALGEBRA, FOR BEGINNERS. 

IN PR EPA RA TION. 



VII. HIGHER ALGEBRA. 

IN PREPARATION. 



Single copies of these books (except the last) will be sent free to teachers of 
mathematics, for their inspection. 



GEORGE W. JONES, Publisher, 



No Agents. 



Ithaca, N. Y. 



DRILL-BOOK 



TEIGONOMETEY 



GEORGE WILLIAM JONES, 

PROFESSOR OF MATHEMATICS IN CORNELL UNIVERSITY. 







•vj 






FIRST EDITION. 



>*. 



FEB- 10 1896, 

ITHACA, N. Y. 
GEORGE W. JONES. 

1896. 



\c\r 



3 



Copyright, 1896, by 
GEORGE WILLIAM JONES. 






Press of J. J. Little & Co. 
Astor Place, New York 



PREFACE. 



In 1881 a treatise ox trigonometry was published under 
the joint authority of Professors Oliver, Wait, and Jones. 
In 1889 this book was rewritten and reissued under the same 
title and by the same authority. In all five editions have been 
printed. 

Professor Oliver died last March, and now that a new edi- 
tion of the book is called for and many changes are proposed, 
it seems better, perhaps fairer towards him, to issue it under 
my single name. It may be regarded, then, both as a new edi- 
tion of the older book, and as itself a new book. 

Among the more important changes are these : 

1. The introduction, at the beginning, of a chapter on the 
right triangle, treating it as the pupil has been accustomed 
to think of it in plane geometry, and without the complex 
notions of directed lines and angles. 

In this chapter he learns, also, how to use tables of trigono- 
metric ratios and logarithms, and he gets some notiou of the 
simpler applications of trigonometry to problems in surveying. 

2. The second chapter, on the general properties of 
plane angles, follows more closely the general lines of the 
old treatise, but it differs widely in details : in particular, it 
makes a much freer use of projections. 

3. The third chapter, on plane triangles, shows a more 
radical departure. The habit of writers on trigonometry 
seems to have been to give broad and general definitions of 
trigonometric ratios, and to prove generally the propositions 
that relate to plane angles, and then, when they come to dis- 
cuss the properties of plane triangles, to forget all they had 
said before, and to fall back on the ratios of positive acute 
angles. 

In the edition of 1889 I tried to make the definitions and 
the proofs general ; but the method then followed never satis- 



IV PREFACE. 

ficcl me, and I sought in vain for light in the many American 
and foreign text-books that I consulted. 

But now, through a happy suggestion of one of my assist- 
ants, Mr. Fowler, I think I have overcome the difficulty. 
That suggestion was to use the exterior angles ; and by 
such use I have been able to make the proofs general and the 
formulas symmetric. So, in space trigonometry, I have been 
able to apply this suggestion to the discussion of the proper- 
ties of triedral angles and spherical triangles with the best 
results. 

4. Greater prominence has been given to the general tri- 
angles. 

5. The proof of De Moivre's formula by aid of imaginaries 
has been left out : I propose to write a book, shortly, on 
higher algebra, and it has seemed to me that there would 
be the best place to discuss the applications of imaginaries to 
trigonometry. 

6. Most of the figures have been redrawn. 

On the other hand, many parts of the older book have been 
included without change, notably the discussion of derivatives 
and series, of directed areas, of astronomy, and of navigation ; 
and for the most part the examples have been taken bodily. 

As to the title of the book, it has seemed to me that the 
word treatise was too large for me ; and as I have meant my 
book primarily for class use, I have called it a drill-book. 

In writing this book, I have been very fortunate in my as- 
sistants. To Mr. Charles S. Fowler and Dr. Virgil Snyder, 
instructors in mathematics in Cornell University, I am deeply 
indebted, both for their valuable suggestions, and for their 
unwearied labors in beating out the text and in preparing the 
questions and examples ; and, for its dress, I am no less 
indebted to my draughtsmen, Mr. John S. Eeid and Mr. Hiram 
• S. Grutsell, instructors in drawing, to my engravers, the 
American Bank Note Company, and to my printers, Messrs. 
J. J. Little & Co. 

George W. Jones. 

Ithaca, N. Y., January 1, 1896. 



SUGGESTIONS TO TEACHERS. 



SUGGESTIONS TO TEACHERS. 



There are many things in this book not meant for beginners. 
Below is a rough list of the chapters and parts of chapters that 
may be taken up at a first reading : the parts omitted are for 
advanced classes. And as to those parts which are included in 
the list, great caution must be taken lest too many examples, 
or too hard ones, be set ; for there are many of them, printed 
in a small space. No one can be expected to work them all, 
and the hardest of them should be reserved for the strongest 
pupils. But the profit comes to the pupil by hard thinking ; 
and the best part of the thinking is in answering the questions. 

Very often more than one figure is used to illustrate a prin- 
ciple : for the most part, the first figure is the simplest, and that 
one should be well understood before the others are looked at. 
Later the other figures may be taken up, and the generality of 
the principle will be felt only when they have all been studied. 

When the reasons are obvious, both theorems and corollaries 
are left without formal demonstration ; but students are ex- 
pected to state the proofs. 

In most cases theorems are given only in formula : it is best 
that these formulae be translated into words. 

In most cases answers to the examples are not given, and the 
student is left to test his own results : the testing is counted 
as not less important than the solution, and the habit of inde- 
pendent thought and self-reliance so cultivated as most valu- 
able of all. 

Only the main lines of the subject are developed in the text : 
collateral matters are outlined in the examples and left for the 
student to work out for himself. 

FOR A FIRST READING. 

I, all, pp. 1-21. 

II, §§ 1-9, 12, pp. 22-53, 58-60. 

III, §§ 1-4, pp. 62-75. 

IV, none. 

V, §§ 1-7, 9-15, pp. 101-130, 134,161. 



VI NEW SIGNS AND WORDS. 

NEW SIGNS AND WORDS. 

Some of the less familiar signs used in this book are these : 
>, larger* than ; ~h, not larger than ; 
<, smaller than ; -£, not smaller than ; 
>, not greater than ; <, not less than ; 
=£, not equal to ; • • •, and so on, meaning the contin- 
uance of a series of terms in the way it has begun ; 
= , approaches, meaning that the value of one expres- 
sion comes very close to that of another, without ab- 
solute equality ; 
=, stands for, or is identical' with. 
The common point of two or more lines or planes is their co- 
point ; the common line of two or more points or planes is their 
co-line ; and the common plane of two or more points or lines 
is their co-plane. The corresponding adjectives are co-pointar, 
co-linear, and co-planar. 

The distinction between larger -smaller inequalities and 
greater-less inequalities is this : the first refers to absolute mag- 
nitude alone, without regard to signs of quality ; the other, in 
common usage, regards both sign and magnitude. 



CONTENTS. 



FOUR-PLACE LOGARITHMS. 
I. THE RIGHT TRIANGLE. 

SECTION PAGE 

1. Trigonometric ratios 1 

2. Trigonometric tables, 8 

3. The solution of right triangles, . 10 

4. Isosceles and oblique triangles, 12 

5. Heights and distances, 14 

6. Compass surveying, 18 

II. GENERAL PROPERTIES OP PLANE ANGLES. 

1. Directed lines, 22 

2. Directed planes and angles, 25 

3. Projections, 31 

4. Trigonometric ratios, 34 

5. Relations of ratios of a single angle, 38 

6. Ratios of related angles, 41 

7. Projection of a broken line, 47 

8. Ratios of the sum, and of the difference, of two angles, ... 48 

9. Ratios of double angles and of half angles, 52 

10. Ratios of the sum of three or more angles and of multiple angles, 54 

11. Inverse functions, 5G 

12. Graphic representation of trigonometric ratios, 58 

III. PLANE TRIANGLES. 

1. The general triangle, 02 

2. General properties of plane triangles, .64 

3. Solution of plane triangles, 08 

4. Sines and tangents of small angles, 74 

5. Directed areas, 70 

0. Inscribed, escribed, and circumscribed circles, 85 



Vlll CONTENTS. 

IV. DERIVATIVES, SEMES, AND TABLES. 

SECTION PAGE 

1. Circular measure of angles, 88 

2. Derivatives of trigonometric ratios, 90 

3. Expansion of trigonometric ratios, 95 

4. Computation of trigonometric ratios, 99 

V. SPACE TRIGONOMETRY. 

1. Directed planes, 104 

2. Diedral angles, 107 

3. Projections, 110 

4. Triedral angles and spherical triangles, 112 

5. General properties of triedral angles, 118 

6. Graphical solution of triedral angles, 121 

7. Four-part formulae, 124 

8. Angles between lines in space, and between planes, . . . .131 

9. Five-part formulae, 134 

10. Six-part formulae, 139 

11. The right triedral, 143 

12. The ideal triedral, 146 

13. Ideal right triangles, 148 

14. Solution of ideal right triangles, 150 

15. Solution of ideal oblique triangles, 156 

16. Relations of plane and spherical triangles, 162 

17. Legendre's theorem, 165 

18. The general spherical triangle, 166 

19. Spherical astronomy, 173 

20. Navigation, 182 



FOUR-PLACE LOGARITHMS. 



FORM OF A LOGARITHM. 

The Logarithm of a number is the exponent of that power to which another number, 
the base, must be raised to give the number first named. The base commonly used is 10 ; 
and as most numbers arc incommensurable powers of 10, a common logarithm, in general, 
consists of an integer, the characteristic \ and an endless decimal, the mantissa. 

If a number be resolved into two factors, of which one is an integer power of 10 and the 
other lies between 1 and 10, then the exponent of 10 is the characteristic, and the logarithm 
of the other factor is the mantissa. The characteristic is positive if the number be larger 
than 1, and negative if it be smaller ; the mantissa is always positive. A negative character- 
istic is indicated by the sign — above it. The logarithms of numbers that differ only by 
the position of the decimal point have different characteristics but the same mantissa. 
E.g. 7770 = 10 3 x 7.77 and log 7770 = 3.8904 ; .0777 = 10" 2 x 7.77, and log .0777 = 2.8904. 

TABLES OF LOGARITHMS. 

The logarithms of any set of consecutive numbers, arranged in a form convenient for 
use, constitute a table of logarithms. Such a table to the base 10 need give only the man- 
tissas ; the characteristics are manifest. This table is arranged upon the common double- 
entry plan; i.e. the mantissa of the logarithm of a three-figure number stands opposite the 
first two figures and under the third figure. The logarithms are given correct to four places. 

TO TAKE OUT THE LOGARITHM OF A NUMBER. 

A three-figure number : Take out the tabular mantissa that lies in line with the first two 
figures of the number and under the third figure ; the characteristic is the exponent of that 
integer power of 10 which lies next below the number. 
E.g. log 677 = 2.8306, log 6.78 = 0.8312, log .0679 = 2.8319, log "676 COO = 5.8299. 

A number of less than three figures : Make the number a three-figure number by annex- 
ing zeros, and follow the rule given above. 
E.g. log 700 = 2.8451, log 7 = 0.8451, log .0071 = 3.8513, log 71 000 = 4.8513. 

A four-figure number : Take out the tabular mantissa of the first three figures, and add 
such part of the difference between this mantissa and the next greater tabular mantissa 
(the tabular difference), as the fourth figure is a part of 10 ; and so for a five-figure number. 
E.g. v log 678 =2.8312 and log 679 = 2.8319, 

.'. log 678.G = 2.8312 + .0007 x 6/10 = 2.8316, log 6.7875=0.8312 + . 0007 x 75/100=0.8317. 

TO TAKE OUT A NUMBER FROM ITS LOGARITHM. 

The mantissa found in the table : Join the figure at the top that lies above the given 
mantissa to the two figures upon the same line at the extreme left ; in this three-figure 
number so place the decimal point that the number shall be next above that power of 10 
whose exponent is the characteristic of the logarithm. 
E.g. log- 1 2.8312 = 678, log-* 0.8451 = 7, log-* 3.8513 = .0071, log-l 5.8513 = 710 0C0. 

The mantissa not found in the table: Takeout the three-figure number of the tabular 
mantissa next less than the given mantissa, and to these three figures join the quotient of 
the difference of these two mantissas by the tabular difference. 
E.g. v log 678 = 2.8312 and log 679 = 2.8319, 

.'. log-i 2.8316 = 678f = 678.6, log-i 2.8317 = .0678f = .06787. 



The use of trigonometric ratios and their logarithms is explained in works on trigonometry. 



LOGARITHMS OF XI' M HERS. 





1 





1 


2 


3 


4 


5 


6 


7 


8 


9 





0000 


0000 


3010 


4771 


6021 


6990 


7782 


8451 


9031 


9542 


1 


0000 


0414 


0792 


1139 


1461 


1761 


2041 


2304 


2553 


2788 


o 


3010 


3222 


3424 


3617 


3802 


3979 


4150 


4314 


4472 


4624 


3 


4771 


4914 


5051 


5185 


5315 


5441 


5563 


5682 


5798 


5911 


4 


6021 


6128 


6232 


6335 


6435 


6532 


6628 


6721 


6812 


6902 


5 


6990 


7076 


7160 


7243 


7324 


7404 


7482 


7559 


7634 


7709 


6 


7782 


7853 


7924 


7993 


8062 


8129 


8195 


8261 


8325 


8388 


7 


8451 


8513 


8573 


8633 


8692 


8751 


8808 


8865 


8921 


8976 


8 


9031 


9085 


9.138 


9191 


9243 


9294 


9345 


9395 


9445 


9494 


9 


9542 


9590 


9638 


9685 


9731 


9777 


9823 


9868 


9912 


9956 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


225 3 


2279 


17 


2304 


2330 


2 3 5 5 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 


3202 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


23 


3617 


3636 


3655 


367 4 


3692 


3711 


3729 


3747 


3766 


3784 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3 927 


3945 


3962 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


2T 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


52 89 


5302 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


36 


5563 


5 5 75 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


56 70 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


40 


G021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


42 


6232 


6243 


6253 


62G3 


6274 


6284 


6294 


G304 


6314 


63 2 5 


43 


6 3 3 5 


63 45 


6355 


6 3 6 5 


6375 


6385 


6395 


6405 


6415 


6425 


44 


6 4 3 5 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


45 


6 532 


6542 


6 5 51 


6561 


0571 


6580 


6590 


6 5 9 9 


6609 


6618 


46 


6628 


663? 


6646 


6 6 56 


6665 


0675 


6684 


6693 


6702 


6712 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6 8 7 5 


6884 


6893 


49 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6 9&1 


50 





1 


2 


3 


4 


5 


6 


7 


8 


9 



LOGARITHMS OF NUMBERS. 



XI 



50 





1 


2 


3 


4 


5 


6 


7 


8 


9 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


57 


7' 5 5 9 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


58 


7634 


7642 


7649 


7657 


7 664 


7672 


7679 


7686 


7694 


7701 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


GO 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


62 


7924 


7931 


7938 


7945 


-7952 


7959 


7966 


7973 


7980 


7987 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


8370 


83 76 


8382 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8 445 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 


8513 


8519 


8525 


8531 


8537 


§543 


8549 


8555 


8561 


8567 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


73 


8633 


8639 


8645 


8651 


86 57 


8663 


8669 


8675 


8681 


8686 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8733 


8739 


8745 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


79 


8976 


8982 


8987 


8993 


89C8 


9004 


9009 


9015 


9020 


9025 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


84 


9243 


9248 


9253 


9258 


92 6 3 


9269 


9274 


9279 


9284 


9289 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489" 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


95 3 8 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 


98 


9312 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 


99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 


100 





1 


2 


3 


4 


5 


6 


7 


8 


9 



Xll 



TRIGONOMETRIC RATIOS. 



ANGLE. 


SINES. 


COSINES. 


TANGENTS. COTANGENTS. 


ANGLE. 




Nat. Log. 


Nat. Log. 


Nat. Log. Log. 


Nat. 




0°00' 


.0000 QO 


1.0000 0.0000 


.0000 QO QO 


GO 


90°00' 


10 


.0029 7.4637 


1.0000 0000 


.0029 7.4637 2.5363 


343.77 


50 


20 


.0058 7648 


1.0000 0000 


.0058 7648 2352 


171.89 


40 


30 


.0087 9408 


1.0000 0000 


.0087 9409 0591 


114.59 


30 


40 


.0116 8.0658 


.9999 0000 


.0116 8.0658 1.9342 


85.940 


20 


50 


.0145 1627 


.9999 0000 


.0145 1627 8373 


68.750 


10 


1°00' 


.0175 8.2419 


.9998 9.9999 


.0175 8.2419 1.7581 


57.290 


89°00' 


10 


.0204 3088 


.9998 9999 


.0204 3089 61)11 


49.104 


50 


20 


.0233 3668 


.9997 9999 


.0233 3669 6331 


42.964 


40 


30 


.0262 4179 


.9997 9999 


.0262 4181 5819 


38.188 


30 


40 


.0291 4637 


.9996 9998 


.0291 4638 5362 


34.368 


20 


50 


.0320 5050 


.9995 9998 


.0320 5053 4947 


31.242 


10 


2°00' 


.0349 8.5428 


.9994 9.9997 


.0349 8.5431 1.4569 


28.636 


88°00 / 


10 


.0378 5776 


.9993 9997 


.0378 5779 4221 


26.432 


50 


20 


.0407 6097 


.9992 9996 


.0407 6101 3899 


24.542 


40 


30 


.0436 6397 


.9990 9996 


.0437 6401 3599 


22.904 


30 


40 


.0465 6677 


.9989 9995 


.0466 6682 3318 


21.470 


20 


5<y 


.0494 6940 


.9988 9995 


.0495 6945 3055 


20.206 


10 


3°00 / 


.0523 8.7188 


.9986 9.9994 


.0524 8.7194 1.2806 


19.081 


87°00' 


10 


.0552 7423 


.9985 9993 


.0553 7429 2571 


18.075 


50 


20 


.0581 7645 


.9983 9993 


.0582 7652 2348 


17.169 


40 


30 


.0610 7857 


.9981 9992 


.0612 7865 2135 


16.350 


30 


40 


.0640 8059 


.9980 9991 


.0641 8067 1933 


15.605 


20 


50 


.0669 8251 


.9978 9990 


.0670 8261 1739 


14.924 


10 


4°00 / 


0698 8.8436 


.9976 9.9989 


.0699 8.8446 1.1554 


14.301 


86°00' 


10 


.0727 8613 


.9974 9989 


.0729 8624 1376 


13.727 


50 


20 


.0756 8783 


.9971 9988 


.0758 8795 1205 


13.197 


40 


30 


.0:85 8946 


.9969 9987 


.0787 8960 1040 


12.706 


30 


40 


.0814 9104 


.9967 9986 


.0816 9118 0882 


12.251 


20 


50 


.0843 9256 


.9964 9985 


.0846 9272 0728 


11.826 


10 


5°00' 


.0S72 8.9403 


.9962 9.9983 


.0875 8.9420 1.0580 


11.430 


85°C0 / 


10 


.0901 9545 


.9959 9982 


.0904 9563 0437 


11.059 


50 


20 


.0929 9682 


.9957 9981 


.0934 9701 0299 


10.712 


40 


30 


.0^58 9816 


.9954 9980 


.0963 9836 0164 


10.385 


30 


40 


.0987 9945 


.9951 9979 


.0992 9966 0034 


10.078 


20 


50 


.1016 9.0070 


.9948 9977 


.1022 9.0093 0.9907 


9.7882 


10 


6 C 00' 


.104 5 9.0192 


.9945 9.9976 


.1051 9.0216 0.9784 


9.5144 


84°00 / 


10 


.1074 0311 


.9942 9975 


.1080 0336 9664 


9.2553 


50 


20 


.1103 0426 


.9939 9973 


.1110 0453 9547 


9.0098 


40 


30 


.1132 0539 


.9936 9972 


.1139 0567 9433 


8.7769 


30 


40 


.1161 0648 


.9932 9971 


.1169 0678 9322 


8.5555 


20 


50 


.1190 0755 


.9929 9969 


.1198 0786 0214 


8.3450 


10 


7°00' 


.1219 9.0859 


.9925 9.9968 


.1228 9.0891 0.91C9 


8.1443 


83°00' 


10 


.1248 0961 


.9922 9966 


.1257 0995 9005 


7.9530 


50 


20 


.1276 1060 


.9918 9964 


.1287 1096 8904 


7.7704 


40 


30 


.1305 1157 


.9914 9963 


.1317 1194 88G6 


7.5958 


30 


40 


.1334 1252 


.9911 9961 


.1346 1291 8709 


7.4287 


20 


50 


.1363 1345 


.9907 9959 


.1376 1385 8615 


7.2687 


10 


8°oo' 


.1392 9.1436 


.9903 9.9958 


.1405 9.1478 0.8522 


7.1154 


82°00 / 


10 


.1421 1525 


.9899 9956 


.1435 1509 8431 


6.9682 


50 


20 


.1449 1612 


.9894 9954 


.1465 1658 8342 


6.8269 


40 


30 


.1478 1697 


.9890 9952 


.1495 1745 8255 


6.6912 


30 


40 


.1507 1781 


.9886 9950 


.1524 1831 8169 


6.5606 


20 


50 


.1536 1863 


.9881 9948 


.1554 1915 8085 


6.4348 


10 


9°00' 


.1564 9.1943 


.9877 9.9946 


.1584 9.1997 0.8003 


6.3138 


81°00' 




Nat. Log. 


Nat. Loir. 


Nat. Log. Log. 


Nat. 




ANGLE. 


COSINES. 


SINES. 


COTANGENTS. TANGENTS. 


ANGLE. 



TRIGONOMETRIC RATIOS. 



Xlll 



ANGLE. 


SIXES. 


COSINES. 


TANGENTS. COTANGENTS. 


ANGLE. 




Nat. Log. 


Nat. Log. 


Nut. Log. Log. 


Nat. 




9°00' 


.1504 9.1943 


.9877 9.9946 


.1584 9.1997 0.8003 


6.3138 


81°00' 


10 


.1593 2022 


.9872 9944 


.1614 2078 7922 


6.1970 


50 


20 


.1622 2100 


.9868 9942 


.1644 2158 7842 


6.0844 


40 


30 


.1650 2176 


.9863 9940 


.1673 2236 7764 


5.9758 


30 


40 


.1679 2251 


.9858 9938 


.1703 2313 7687 


5.8708 


20 


50 


.1708 2324 


.9853 9936 


.1733 2389 7611 


5.7694 


10 


10°00' 


.1736 9.2397 


.9848 9.9934 


.1763 9.24G3 0.7537 


5.6713 


80°00' 


10 


.1765 2468 


.9843 9931 


.1793 2536 7464 


5.5764 


50 


20 


.1794 2538 


.9838 9929 


.1823 2609 7391 


5.4845 


40 


30 


.1822 2606 


.9833 9927 


.1853 2680 7320 


5.3955 


30 


40 


.1851 2674 


.9827 9924 


.1883 2750 7250 


5.3093 


20 


50 


.1880 2740 


.9822 9922 


.1914 2819 7181 


5.2257 


10 


11°00' 


.1908 9.2806 


.9816 9.9919 


.1944 9.2887 0.7113 


5.1446 


79°00' 


10 


.1937 2870 


.9811 9917 


.1974 2953 7047 


5.0658 


50 


20 


.1965 2934 


.9805 9914 


.2004 3020 6980 


4.9894 


40 


30 


.1994 2997 


.9799 9912 


.2035 3085 6915 


4.9152 


30 


40 


.2022 3058 


.9793 9909 


.2065 3149 6851 


4.8430 


20 


50 


.2051 3119 


.9787 9907 


.2095 3212 6788 


4.7729 


10 


12°00' 


.2079 9.3179 


.9781 9.9904 


.2126 9.3275 0.6725 


4.7046 


78°00 / 


10 


.2108 3238 


.9775 9901 


.2156 3336 6664 


4.6382 


50 


20 


.2136 3296 


.9769 9899 


.2186 3397 6603 


4.5736 


40 


30 


.2164 3353 


.9763 9896 


.2217 3458 6542 


4.5107 


30 


40 


.2193 3410 


.9757 9893 


.2247 3517 6483 


4.4494 


20 


50 


.2221 3466 


.9750 9890 


.2278 3576 6424 


4.3897 


10 


13°00' 


.2250 9.3521 


.9744 9.9887 


.2309 9.3634 0.6366 


4.3315 


77°00' 


10 


.2278 3575 


.9737 9884 


.2339 3691 6309 


4.2747 


50 


20 


.2306 3629 


.9730 9881 


.2370 3748 6252 


4.2193 


40 


30 


.2334 3682 


.9724 9878 


.2401 3804 6196 


4.1653 


30 


40 


.2363 3734 


.971? 9875 


.2432 3859 6141 


4.1126 


20 


50 


.2391 3786 


.9710 9872 


.2462 3914 6086 


4.0611 


10 


14°00' 


.2419 9.3837 


.9703 9.9869 


.2493 9.3968 0.6032 


4.0108 


76°00> 


10 


.2447 3887 


.9696 9866 


.2524 4021 5979 


3.9617 


50 


20 


.2476 3937 


.9689 9863 


.2555 4074 5926 


3.9136 


40 


30 


.2504 3986 


.9681 9859 


.2586 4127 5873 


3.8667 


30 


40 


.2532 4035 


.9674 9856 


.2617 4178 5822 


3.8208 


20 


50 


.2560 4083 


.9667 9853 


.2648 4230 5770 


3.7760 


10 


15°00' 


.2588 9.4130 


.9659 9.9849 


.2679 9.4281 0.5719 


3.7321- 


75°00 / 


10 


.2616 4177 


.9652 9846 


.2711 4331 5669 


3.6891 


50 


20 


.2644 4223 


.9644 9843 


.2742 4381 5619 


3.6470 


40 


30 


.2672 4269 


.9636 9839 


.2773 4430 5570 


3.6059 


30 


40 


.2700 4314 


.9628 9836 


.2805 4479 5521 


3.5656 


20 


50 


.2728 4359 


.9621 9832 


.2836 4527 5473 


3.5261 


10 


16°00' 


.2756 9.4403 


.9613 9.9828 


.2867 9.4575 0.5425 


3.4874 


74°00' 


10 


.2784 4447 


.9605 9825 


.2899 4622 5378 


3.4495 


50 


20 


.2812 4491 


.9596 9821 


.2931 4669 5331 


3.4124 


40 


30 


.2840 4533 


.9588 9817 


.2962 4716 5284 


3.3759 


30 


40 


.2868 4576 


.9580 9814 


.2994 4762 5238 


3.3402 


20 


50 


.2896 4618 


.9572 9810 


.3026 4808 5192 


3.3052 


10 


17°00' 


.2924 9.4659 


.9563 9.9806 


.3057 9.4853 0.5147 


3.2709 


73°00' 


10 


.2952 4700 


.9555 9802 


.3089 4898 5102 


3.2371 


50 


20 


.2979 4741 


.9546 9798 


.3121 4943 5057 


3.2041 


40 


30 


.3007 4781 


.9537 9794 


.3153 4987 5013 


3.1716 


30 


40 


.3035 4821 


.9528 9790 


.3185 5031 4969 


3.1397 


20 


50 


.3062 4861 


.9520 9786 


.3217 5075 4925 


3.1084 


10 


18°00' 


.3090 9.4900 


.9511 9.9782 


.3249 9.5118 0.4882 


3.0777 


72°00' 




Nat. Log. 


Nat. Log. 


Nat. Log. Log. 


Nat. 




ANGLE. 


COSINES. 


8INES. 


COTANGENTS. TANGENTS. 


ANGLE. 



XIV 



TRIGONOMETRIC RATIOS. 





ANGLE. 


SINES. 


COSINES. 


TANGENTS. COTANGEN 'S. 


ANGLE. 


Nat. Log. 


Nat. Log. 


Nat, Log. Log. 


Nat. 


18°00' 


.3090 9.4900 


.9511 9.9782 


.3249 9.5118 0.4882 


3.0777 


72°00' 


10 


.3118 4939 


.9502 9778 


.3281 5161 4839 


3.0475 


50 


20 


.3145 4977 


.9492 9774 


.3314 5203 4797 


3.0178 


40 


30 


.3173 5015 


.9483 .9770 


.3346 5245 4755 


2.9887 


3 


40 


.3201 5052 


.9474 9765 


.3378 5287 4713 


2.9600 


20 


50 


.3228 5090 


.9465 9761 


.3411 5329 4671 


2.9319 


10 


19°00' 


.3256 9.5126 


.9455 9.9757 


.3443 9.5370 0.4630 


2.9042 


71°00' 


10 


.3283 5163 


.9446 9752 


.3476 5411 4589 


2.8770 


50 


20 


.3311 5199 


.9436 9748 


.3508 5451 4549 


2.8502 


40 


30 


.3338 5235 


.9426 9743 


.3541 5491 4509 


2.8239 


30 


40 


.3365 5270 


.9417 9739 


.3574 5531 4469 


2.7980 


20 


50 


.3393 5306 


.9407 9734 


.3607 5571 4429 


2.7725 


10 


20°00' 


•3420 9.5341 


.9397 9.9730 


.3640 9.5611 0.4389 


2.7475 


70°00 / 


10 


.3448 5375 


.9387 9725 


.3673 5650 4350 


2.7228 


50 


20 


.3475 5409 


.9377 9721 


.3706 5689 4311 


2.6985 


40 


30 


.3502 5443 


.9367 9716 


.3739 5727 4273 


2.6746 


30 


40 


.3529 5477 


.9356 9711 


.3772 5766 4234 


2.6511 


20 


50 


.3557 5510 


.9346 9706 


.3805 5804 4196 


2.6279 


10 


21°00' 


.3584 9.5543 


.9336 9.9702 


.3839 9.5842 0.4158 


2.6051 


69°oo' 


10 


.3611 5576 


.9325 9697 


.3872 5879 4121 


2.5826 


50 


20 


.3638 5609 


.9315 9692 


.3906 5917 4083 


2.5605 


40 


30 


.3665 5641 


.9304 9687 


.3939 5954 4046 


2.5386 


30 


40 


.3692 5673 


.9293 9682 


.3973 5991 4009 


2.5172 


20 


50 


.3719 5704 


.9283 9677 


.4006 6028 3972 


2.4960 


10 


22°00' 


.3746 9.5736 


.9272 9.9672 


.4040 9.6064 0.3936 


2.4751 


6s°oo' 


10 


.3773 5767 


.9261 9667 


.4074 6100 3900 


2.454 5 


50 


20 


.3800 5798 


.9250 9661 


.4108 6136 3864 


2.4342 


40 


30 


.3827 5828 


.9239 9656 


.4142 6172 3828 


2.4142 


30 


40 


.3854 5859 


.9228 9651 


.4176 6208 3792 


2.3945 


20 


50 


.3881 5889 


.9216 9646 


.4210 6243 3757 


2.3750 


10 


23*00' 


.3907 9.5919 


.9205 9.9640 


.4245 9.6279 0.3721 


2.3559 


67°00 / 


10 


.3934 5948 


.9194 9635 


.4279 6314 3686 


2.3369 


50 


20 


.3961 5978 


.9182 9629 


.4314 6348 3652 


2.3183 


40 


30 


.3987 6007 


.9171 9624 


.4348 6383 3617 


2.2998 


30 


40 


.4014 6036 


.9159 9618 


.4383 6417 3583 


2.2817 


20 


50 


.4041 "6065 


.9147 9613 


.4417 6452 3548 


2.2637 


10 


;24°00' 


.4067 9.6093 


.9135 9.9607 


.4452 9.6486 0.3514 


2.2460 


66°00' 


10 


.4094 6121 


.9124 9602 


.4487 6520 3480 


2.2286 


50 


20 


.4120 6149 


.9112 9596 


.4522 6553 3447 


2.2113 


40 


30 


.4147 6177 


.9100 9590 


.4557 6587 3413 


2.1943 


30 


40 


.4173 6205 


.9088 i)584 


.4592 6620 3380 


2.177 5 


20 


50 


.4200 6232 


.9075 9579 


.4628 6654 3346 


2.1609 


10 


25°00' 


.4226 9.6259 


.9063 9.9573 


.4663 9.6687 0.3313 


2.1445 


65°00' 


10 


.4253 6286 


.9051 9567 


.4699 6720 3280 


2.1283 


50 


20 


.4279 6313 


.9038 9561 


.4734 6752 3248 


2.1 123 


40 


30 


.4305 6340 


.9026 9555 


.4770 6785 3215 


2.0965 


30 


40 


.4331 6366 


.9013 9549 


.4806 6817 3183 


2.0fe0 9 


20 


50 


.4358 6392 


.9001 9543 


.4841 6850 3150 


2.0655 


10 


26°00' 


.4384 9.6418 


.8988 9.9537 


.4877 9.6882 0.3118 


2.0503 


64° 00' 


10 


.4410 6444 


.8975 9530 


.4913 6914 3086 


2.0353 


50 


20 


.4436 6470 


.8962 9524 


.4950 6946 3054 


2.0204 


40 


30 


.4462 6495 


.8949 9518 


.4986 6977 3023 


2.005 7 


30 


40 


.4488 6521 


.8936 9512 


.5022 7009 2991 


1.9912 


20 


50 


.4514 6546 


.8923 9505 


.5059 7040 2960 


1.9768 


10 


27°00' 


.4540 9.6570 


.8910 9.9499 


.5095 9.7072 0.2928 


1.9626 


63° 00' 




Nat. Log. 


Nat. Log. 


Nat. Log. Log. 


Nat. 

5ENTS. 




ANGLE. 


COSINES. 


SINES. 


COTANGENTS. TAN< 


ANGLE. 



TRIGONOMETRIC RATIOS. 



XV 





ANGLE. 


SINES. 


COSINES. 


TANGENTS. COTANGENTS. 


ANGLE. 




Nat. Log. 


Nat. Log. 


Nat. Log. Log. 


Nat. 




27°00 / 


.4540 9.6570 


.8910 9.9499 


.5095 9.7072 0.2928 


1.9626 


63*00' 


10 


.4566 6595 


.8897 9492 


.5132 7103 2897 


1.9486 


50 . 


20 


.4592 6620 


.8884 9486 


.5169 7134 2866 


1.9347 


40 


30 


.4617 6644 


.8870 9479 


.5206 7165 2835 


1.9210 


30 


40 


.4643 6668 


.8857 9473 


.5243 7196 2804 


1.9074 


20 


50 


.4669 6692 


.8843 9466 


.5280 7226 2774 


1.8940 


10 


28°00' 


.4695 9.6716 


.8829 9.9459 


.5317 9.7257 0.2743 


1.8807 


62°00 / 


10 


.4720 6740 


.8816 9453 


.5354 7287 2713 


1.8676 


50 


20 


.4746 6763 


.8802 9446 


.5392 7317 2683 


1.8546 


40 


30 


.4772 6787 


.8788 9439 


.5430 7348 2652 


1.8418 


30 


40 


.4797 6810 


.8774 9432 


.5467 7378 2622 


1.8291 


20 


50 


.4823 6833 


.8760 9425 


.5505 7408 2592 


1.8165 


10 


29°00' 


.4848 9.6856 


.8746 9.9418 


.5543 9.7438 0.2562 


1.8040 


61°00' 


10 


.4874 6878 


.8732 9411 


.5581 7467 2533 


1.7917 


50 


20 


.4899 6901 


.8718 9404 


.5619 7497 2503 


1.7796 


40 


30 


.4924 6923 


.8704 9397 


.5658 7526 2474 


1.7675 


30 


40 


.4950 6946 


.8689 9390 


.5696 7556 2444 


1.7556 


20 


50 


.4975 6968 


.8675 9383 


.5735 7585 2415 


1.7437 


10 


30°00' 


.5000 9.6990 


.8660 9.9375 


.5774 9.7614 0.2386 


1.7321 


60°00' 


10 


.5025 7012 


.8646 9368 


.5812 7644 2356 


1.7205 


50 


20 


.5050 7033 


.8631 9361 


.5851 7673 2327 


1.7090 


40 


30 


.5075 7055 


.8616 9353 


.5890 7701 2299 


1.6977 


30 


40 


.5100 7076 


.8601 9346 


.5930 7730 2270 


1.6884 


20 


50 


.5125 7097 


.8587 9338 


.5969 7759 2241 


1.6753 


10 


3i°oo' 


.5150 9.7118 


.8572 9.9331 


.6009 9.7788 0.2212 


1.6643 


59°00' 


10 


.5175 7139 


.8557 9323 


.6048 7816 2184 


1.6534 


50 


20 


.5200 7160 


.8542 9315 


.6088 7845 2155 


1.6426 


40 


30 


.5225 7181 


.8526 9308 


.6128 7873 2127 


1.6319 


30 


40 


.5250 7201 


.8511 9300 


.6168 7902 2098 


1.6212 


20 


50 


.5275 7222 


.8496 9292 


.6208 7930 2070 


1.6107 


10 


32°00' 


.5299 9.7242 


.8480 9.9284 


.6249 9.7958 0.2042 


1.6003 


58°00 / 


10 


.5324 7262 


.8465 9276 


.6289 7986 2014 


1.5900 


50 


- 20 


.5348 7282 


.8450 9268 


.6330 8014 1986 


1.5798 


40 


30 


.5373 7302 


.8434 9260 


.6371 8042 1958 


1.5697 


30 


40 


.5398 7322 


.8418 9252 


.6412 8070 1930 


1.5597 


20 


50 


.5422 7342 


.8403 9244 


.6453 8097 1903 


1.5497 


10 


33°00' 


.5446 9.7361 


.8387 9.9236 


.6494 9.8125 0.1875 


1.5399 


57°00' 


10 


.5471 7380 


.8371 9228 


.6536 8153 1847 


1.5301 


50 


20 


.5495 7400 


.8355 9219 


.6577 8180 1820 


1.5204 


40 


30 


.5519 7419 


.8339 9211 


.6619 8208 1792 


1.5108 


30 


40 


.5544 7438 


.8323 9203 


.6661 8235 1765 


1.5013 


20 


50 


.5568 7457 


.8307 9194 


.6703 8263 1737 


1.4919 


10 


34°00' 


.5592 9.7476 


.8290 9.9186 


.6745 9.8290 0.1710 


1.4826 


56°00 / 


10 


.5616 7494 


.8274 9177 


.6787 8317 1683 


1.4733 


50 


20 


.5640 7513 


.8258 9169 


.6830 8344 1656 


1.4641 


40 


30 


.5664 7531 


.8241 9160 


.6873 8371 1629 


1.4550 


30 


40 


.5688 7550 


.8225 9151 


.6916 8398 1602 


1.4460 


20 


50 


.5712 7568 


.8208 9142 


.6959 8425 1575 


1.4370 


10 


35°00' 


.5736 9.7586 


.8192 9.9134 


.7002 9.8452 0.1548 


1.4281 


55°00' 


10 


.5760 7604 


.8175 9125 


.7046 8479 1521 


1.4193 


50 


20 


.5783 7622 


.8158 9116 


.7089 8506 1494 


1.4106 


40 


30 


.5807 7640 


.8141 9107 


.7133 8533 1467 


1.4019 


30 


40 


.5831 7657 


.8124 9098 


.7177 8559 1441 


1.3934 


20 


50 


.5854 7675 


.8107 9089 


.7221 8586 1414 


1.3848 


10 


36°00' 


.5878 9.7692 


.8090 9.9080 


.7265 9.8613 0.1387 


1.3764 


54°00' 




Nat. Log. 


Nat. Log. 


Nat. Log. Log. 


Nat. 




ANGLE. 


COSINES. 


SINES. 


COTANGENTS. TANGENTS. 


ANGLE. 





XVI 



TRIGONOMETRIC RATIOS. 



ANGLE. 


SINES. 


COSINES. 


TANGENTS. COTANGENTS. 


ANGLE. 




Nat. Log. 


Nat. Log. 


Nat. Log. Log. 


Nat. 




36°00' 


.5878 9.7692 


.8090 9.9080 


.7265 9.8613 0.1387 


1.3764 


54°00 / 


10 


.5901 7710 


.8073 9070 


.7310 8639 1361 


1.3680 


5 


20 


.5925 7727 


.8056 9061 


.7355 8666 1334 


1.3597 


40 


30 


.5948 7744 


.8039 9052 


.7400 8692 1308 


1.3514 


30 


40 


.5972 7761 


.8021 9042 


.7445 8718 1282 


1.3432 


20 


50 


.5995 7778 


.8004 9033 


.7490 8745 1255 


1.3351 


10 


37°00 / 


.6018 9.7795 


.7986 9.9023 


.7536 9.8771 0.1229 


1.3270 


53°oo' 


10 


.6041 7811 


.7969 9014 


.7581 8797 1203 


1.3190 


50 


20 


.6065 7828 


.7951 9004 


.7627 8824 1176 


1.3111 


40 


30 


.6088 7844 


.7934 8995 


.7673 8850 1150 


1.3032 


30 


40 


.6111 7861 


.7916 8985 


.7720 8876 1124 


1.2954 


20 


50 


.6134 7877 


.7898 8975 


.7766 8902 1098 


1.2876 


10 


38°00 / 


.6157 9.7893 


.7880 9.8965 


.7813 9.8928 0.1072 


1.2799 


52°00 / 


10 


.6180 7910 


.7862 8955 


.7860 8954 1046 


1.2723 


50 


20 


.6202 7926 


.7844 8945 


.7907 8980 1020 


1.2647 


40 


30 


.6225 7941 


.7826 8935 


.7954 9006 0994 


1.2572 


30 


40 


.6248 7957 


.7808 8925 


.8002 9032 0968 


1.2497 


20 


50 


.6271 7973 


.7790 8915 


.8050 9058 0942 


1.2423 


10 


39°00' 


.6293 9.7989 


.7771 9.8905 


.8098 9.9084 0.0916 


1.2349 


5i°oo' 


10 


.6316 8004 


.7753 8895 


.8146 9110 0890 


1.2276 


50 


20 


.6338 8020 


.7735 8884 


.8195 9135 0865 


1.2203 


40 


30 


.6361 8035 


.7716 8874 


.8243 9161 0839 


1.2131 


30 


40 


.6383 8050 


.7698 8864 


.8292 9187 0813 


1.2059 


20 


• 50 


.6406 8066 


.7679 8853 


.8342 9212 0788 


1.1988 


10 


40°00' 


.6428 9.8081 


.7660 9.8843 


.8391 9.9238 0.0762 


1.1918 


50°oo' 


10 


.6450 8096 


.7642 8832 


.8441 9264 0736 


1.1847 


50 


20 


.6472 8111 


.7623 8821 


.8491 9289 0711 


1.1778 


40 


30 


.6494 8125 


.7604 8810 


.8541 9315 0685 


1.1708 


30 


40 


.6517 8140 


.7585 8800 


.8591 9341 0659 


1.1640 


20 


50 


.6539 8155 


.7566 8789 


.8642 9366 0634 


1.1571 


10 


41°00 / 


.6561 9.8169 


.7547 9.8778 


.8693 9.9392 0.0608 


1.1504 


49°00 / 


10 


.6583 8184 


.7528 8767 


.8744 9417 0583 


1.1436 


50 


20 


.6604 8198 


.7509 8756 


.8796 9443 0557 


1.1369 


40 


30 


.6626 8213 


.7490 8745 


.8847 9468 0532 


1.1303 


30 


40 


.6648 8227 


.7470 8733 


.8899 9494 0506 


1.1237 


20 


50 


.6670 8241 


.7451 8722 


.8952 9519 0481 


1.1171 


10 


42°00' 


.6691 9.8255 


.7431 9.8711 


.9004 9.9544 0.0456 


1.1106 


48°00 / 


10 


.6713 8269 


.7412 8699 


.9057 9570 0430 


1.1041 


50 


20 


.6734 8283 


.7392 8688 


.9110 9595 0405 


1.0977 


40 


30 


.6756 8297 


.7373 8676 


.9163 9621 0379 


1.0913 


30 


40 


.6777 8311 


.7353 8665 


.9217 9646 0354 


1.0850 


20 


50 


.6799 8324 


.7333 8653 


.9271 9671 0329 


1.0786 


10 


43°00 / 


.6820 9.8338 


.7314 9.8641 


.9325 9.9697 0.0303 


1.0724 


47°00' 


10 


.6841 8351 


.7294 8699 


.9380 9722 0278 


1.0661 


50 


20 


.6862 8365 


.7274 8618 


.9435 9747 0253 


1.0599 


40 


30 


.6884 8378 


.7254 8606 


.9490 9772 0228 


1.0538 


30 


40 


.6905 . 8391 


.7234 8594 


.9545 9798 0202 


1.0477 


20 


50 


.6926 8405 


.7214 8582 


.9601 9823 0177 


1.0416 


10 


44°00' 


.6947 9.8418 


.7193 9.8569 


.9657 9.9848 0.0152 


1.0355 


46°00 / 


10 


.6967 8431 


.7173 8557 


.9713 9874 0126 


1.0295 


50 


20 


.6988 8444 


.7153 8545 


.9770 9899 0101 


1.0235 


40 


30 


.7009 8457 


.7133 8532 


.9827 9924 0076 


1.0176 


30 


40 


.7030 8469 


.7112 8520 


.9884 9949 0051 


1.0117 


20 


50 


.7050 8482 


.7092 8507 


.9942 9975 0025 


1.0058 


10 


45°00 / 


.7071 9.8495 


.7071 9.8495 


l.oooo o.oaoo o.oooo 


1.0000 


45°00 / 




Nat. Log. 


Nat. Log. 


Nat. Log. Log. 


Nat. 




ANGLE. 


COSINES. 


SINES. 


COTANGENTS. TANGENTS. 


ANGLE. 



TRIGONOMETRY. 

Trigonometry is that branch of mathematics which treats 
of the numerical relations of angles and triangles. It is 
essentially algebraic in character, bat is founded on geometry. 

I. THE EIGHT TRIANGLE. 



§1. TRIGONOMETRIC RATIOS. 

Theor. 1. If from a point in one side of an acute angle a 
perpendicular fall on the other side, then, in the right trian- 
gle so formed, the ratio of the side opposite the angle to the 
hypotenuse is the same, whatever point be taken / and so for 
that of the adjacent side to the hypotenuse, for that of the 
opposite side to the adjacent side, and for the reciprocals of 
these three ratios. 

For let A be any acute angle ; B, b' • • • points on either bound- 
ing line ; a, a' • • • perpendiculars from b, b' • • • to the 
other line at c, c'« • • ; b, b' > • • the lines AC, Ac'- • •; and 
c, c • • • the lines ab, ab' • • • ; 




thenvthe right triangles abo, ab'c'- • • are similar, 

.*. the ratios a/c, a'/c'- • ■ are all equal ; 
and so for the other ratios b/c, b'/c' • • • , a/b, a'/b' • 
b/a, b'/a'- ■ ., c/b, c'/b'. ■ -, c/a, c'/a r > • •. 



2 



THE RIGHT TRIAXGLE. 



[I> 



But if an angle be taken greater or less than A, the tri- 
angles so formed are not similar to these, and the ratios are 
different from those for the angle A. 

For this reason an acute angle has its six ratios distinct 
from the ratios of every other angle, and if one of the ratios 
be given the angle can be constructed. These ratios are the 
six principal trigonometric functions of an angle, and they 
are named as follows : 

opposite side to hypotenuse, the sine of the angle, 
adjacent side to hypotenuse, the cosine, 
opposite side to adjacent side, the tangent, 
adjacent side to opposite side, the cotangent, 
hypotenuse to adjacent side, the secant, 
hypotenuse to opposite side, the cosecant. 
When written before the name of the angle, the words 
sine, cosine, tangent, cotangent, secant, cosecant may be 
abbreviated to sin, cos, tan, cot, sec, esc. Standing alone, 
the abbreviations have no meaning. 

If abc be any right triangle with c the right angle, a the 
side opposite the acute angle A, b the side opposite the acute 
angle b, and c the hypotenuse, then the six ratios of each of 
the acute angles may be expressed in terms of the three sides 
of the triangle, as below. 






C 


A 




b 










b/ 




^\a 


s 





sin a = a/c, esc A = c/a, and sin b = b/c, esc b = c/b, 
cos a = b/c, sec a = c/b, cos b = a/c, sec b = c/a, 

tan a = a/b, cot A = b/a, tan b = b/a, cot b = a/b. 

Note. In the discussion of the right triangle that follows, 
the triangle is always lettered as in the figures above ; i.e., 



§1.] TRIGONOMETRIC RATIOS. 3 

with c for the right angle, c for the hypotenuse, A, b for the 
acute angles, a, b for the sides opposite A, b. 

The expression sin -1 a/c means the angle whose sine is 
a/c; cos -1 b/c> the angle whose cosine is i/c; tan -1 a/b, the 
angle whose tangent is a/b, and so for the other ratios. They 
are read : the anti-sine of a/c, the anti-cosine oib/c, the anti- 
tangent of a/b, and so on. 

E.g. if sin A — \, ifcosB = f, if tan f = 6, ifcotx=V3, 
then A = sin-H, B = cos- 1 |, F = tan _1 6, x = cot- l V3. 

The index _1 is to be carefully distinguished from the nega- 
tive exponent ; it is analogous to that in the expression log -1 2, 
which is read the ant i -logarithm of 2 and means the number 
whose logarithm is 2. 

The positive powers of the trigonometric ratios are com- 
monly written in the form sin 2 A, cos 3 b, instead of (sin a) 2 , 
(cos b) 3 ; but their reciprocals are written in the form of 
fractions, or with the exponent without the bracket. 
E.g. 1/sinA, or (sin a) -1 , not sin -1 a ; l/cos 2 B, or (cosb) -2 . 

QUESTIONS. 

1. If the sides a, b, c, of a right triangle abc be 3 feet, 4 
feet, 5 feet, what are the six ratios of a and of b? if the 
sides be 3 yards, 4 yards, 5 yards ? if 3 miles, 4 miles, 5 miles ? 

2. In a right triangle abc the sides a, c are 12 yards and 13 
yards : find b and the six ratios of A and of B. 

So, if a, b be 12 feet and 5 feet, find c and the six ratios. 

3. Construct the right triangle abc with the hypotenuse c 
5 feet, and a side a 3 feet. What is the sine of the angle A ? 

From this construct a right triangle abc if sinA=f. 
So, if cos a = f, if tan A = |, if cot a = 4, if sec a = f. 

4. Construct sin -1 -J-, cos -1 f-, tan -1 |, cot -1 f, sec -1 |. 

5. Find the six ratios of one of the acute angles of a right 
isosceles triangle. 

6. Draw a perpendicular from the vertex to the base of an 
equilateral triangle, and find the six ratios of the acute 
angles of the right triangles so formed. 



4 THE RIGHT TRIANGLE. [I, THS. 

7. In a right triangle abc, let the hypotenuse c be 12 feet 
and the angle A be 30°: find the sides a> b, given sin 30° = .5, 
cos 30° = .866, nearly. 

8. In a right triangle abc, let the side a be 12 yards and 
the angle A be 35° : find the sides b, c, given sin 35° = .574, 
tan 35° = .7. 

9. In a right triangle abc, let the side b be 12 miles and 
the angle Abe 40°: find the sides c> a, given cos 40° = .766, 
cot 40° = 1.192. 

10. In a right triangle abc, let the hypotenuse c be 12 feet 
and the side a be 8.484 feet : find the side b and the angle A, 
given sin 45° = .707. 

11. In a right triangle abc, let the side a be 12 yards and 
the side b be 9.948 yards : find the side c and the angle A, given 
sin 50° = .766, tan 50° = 1.192. 

12. In a right triangle abc, let the side b be 12 miles and 
the hypotenuse c be 20.9 miles : find the side a and the angles, 
given cos 55° = .574, tan 55° = 1.428. 

13. In a right triangle abc, let the side a be 12 metres 
and the hypotenuse c be 33J metres : find the side b and the 
angles, given sin 21° 6' = .36, cos 21° 6' = .933. 

Verify the work by showing that a 2 + 5 2 = c 2 . 

14. Draw two right triangles abc, a'b'c', having A larger 
than a', and show which of the ratios of A are larger, and 
which are smaller, than the like-named ratios of a'. 

15. Draw a right triangle having an acute angle less than 
half a right angle, and show which of the ratios of that angle 
are larger than unity, and which are smaller. 

16. Draw a right triangle having one acute angle very small, 
and show which of the ratios of this angle are very small, 
which are very large, and which are near unity. 

As the angle is made smaller and smaller, approaching zero, 
to what do these ratios approach ? 

So, what are the ratios of the other acute angle, which is 
very near a right angle ? 



2-4, §1.] 



TRIGONOMETRIC RATIOS. 



Theor. 2. If a be any acute angle, then: 

sin A. • esc A = 1, cos a • sec a = 1, tan a • CO*? A == 1. 




For, from any point b of either side of the angle, let fall a 
perpendicular bc upon the other side, as in theor. 1; 
thenvin the right triangle abc so formed, 

sin A = a/c, csc a = <?/#, [df. 

.*. sin a • esc a = a/c-c/a=l. q.e.d. 

So, '/cos A — b/c, secA — c/b, [df. 

.\ cos a • sec A — b/c • c/b = 1. q.e.d. 

So, vtan A — a/b, cotA = b/a, [df. 

.-. tan a • cot a = a/b • b/a = L q.e.d. 

Theor. 3. If a b.e any acute angle, then: 
sin A/ 'cos a — tan A, cos A /sin A — cot A. 
For v in the right triangle abc, formed as in theor. 1, 

sin A = a/c, cos A = b/c, tan a = a/b, cot a = b/a, [df 
.'. sin a/cos a = a/c : b/c = a/b = tariA, 
and cos A/sin A = b/c : a/c = b/a — cot a. q. e. d. 

Theor. 4. If a be any acute angle, then : 

sin 2 a + c$s 2 A = 1, J. + tan 2 a - sec 2 a, 1 + cot 2 A = esc 2 A. 
For v in the right triangle abc, a 2 -\-b 2 — c 2 , 

.\a 2 /c 2 + b 2 /c 2 = l; [div. by c\ 

and v sin A = a/c, cos A = b/c> [df . 

.*. sin 2 A + COS 2 A=l. Q.E.D. 

So, a 2 /b 2 + 1 = c 2 /b 2 ; [div. by b\ 

and y tan A = a/b, sec A = c/b 9 [df . 

.\ l+tan 2 A = sec 2 A. Q.E.D. 

So, 1 + b 2 /a 2 = c 2 /a 2 ; [div. by a 2 . 
and y cot a = b/a, esc A == c/a, 

■\ 1 + COt 2 A = CSC 2 A. Q. E. D. 



THE RIGHT TRIANGLE. 



[i, 



Theor, 5. If a be any acute angle and B the complement 
of a, then : sin a — cos B, tan a = cut b, sec\ = esc B. 
a a b 



B 

For 
and 

So, 




• in the right triangle abc, formed as in theor. 1, A. B 
are complementary acute angles, 
md 



." sin k. — a/Cy 

\ sin a = cos B. 

.' tanA = « b. 

.'. taiiA = cot B. 

Soj v secA = c/b, 
.'. sec A — CSC B. 



cosB = r/ c, 



and cotB = tf/#, 



[df. 



Q.E.D. 



[df. 



Q.E.D. 



ana 



esc B = r b. 



[df. 



Q.E.D. 



Xote. If the sine, tangent, and secant of an angle be called 
its direct ratios, and the cosine, cotangent, and cosecant the 
co-ratios, theor. 5 may be stated as follows : the direct ratios 
of an angle are the co-ratios of its complement. 

The words cosine, cotangent, and cosecant are but abbre- 
viated forms for complement-sine, complement-tangent, and 
complement-secant : i.e. for sine of complement, tangent of 
complement, and secant of complement. 

QUESTIONS. 

1. Translate the equation sin*A + cos*A=l into words, 
and express its meaning as a theorem. 

Solve this equation in turn for sin a and cos a, and trans- 
late the resulting equations into theorems. 

2. Translate the equation sec 2 A = l+tan 2 A into words, 
and express its meaning as a theorem. 

Solve this equation in turn for sec A and tan A, and trans- 
late the resulting equations into theorems. 
So, the equation csc 2 a = 1 -f-cot 2 A. 



5, §1.] TRIGONOMETRIC RATIOS. 7 

3. Show that 

sin A = tan a • cos A = tan A/sec a = cos A/cot a, 
esc A = sec A/tan a = cot a/cos a = cot a • sec A, 
cos A = cot a • sin A = cot a/csc a = sin A/tan a, 
sec A = esc A/cot a = tan A/sin A = tan a • esc A, 
tan a = sin A • sec A = sec a/csc a, 

COt A = COS A • CSC A = CSC A/sec A. 

Translate these equations into theorems. 

4. Show that 

sin A = tan a/ V (tan 2 a + 1) = V (sec 2 A — l)/sec A, 
csca= V(tan 2 A + l)/tan A = sec A/V(sec 2 A — 1), 

COS A = COt A./V(C0t 2 A + 1) = V(CSC 2 A — 1)/CSC A, 

sec A = V (cot 2 A + l)/cot A = csc a/V (csc 2 a — 1), 
tanA = sin a/V(1 — sin 2 A) = V(l — cos 2 a)/cos a, 
cot a = V(l — sin 2 A)/sin a = cos a/V (1 — cos 2 a). 
Translate 'these equations into theorems. 

5. If the hypotenuse c of a right triangle abc have unit 
length, show that the two legs a, b, have the lengths sin A, 
V(l — sin 2 A), and thence find the values of tan a, cot a, 
sec A, in terms of sin a. 

So, show that the two legs a, i, have the lengths 
V(l — cos 2 a), cos a, and thence find the values of tan a, 
cot a, csc a, in terms of cos a. 

6. If the leg ~b of a right triangle abc have unit length, 
show that the leg a and hypotenuse c have the lengths tan a, 
V(tan 2 A + l), and thence find the values of sin a, cos a, 
sec a, csc a. in terms of tan a. 

7. With the values of the ratios of the angles 30°, 45°, 60°, 
as found in examples 5, 6, page 3, find the values of A from 
the equations : tan a + cot a =2, sin A -h cos A = */2, 

sin A • secA= \/3, cot a = 2 cos A. 

8. Find the other ratios of a 

if sin A =f, if cos a = f, if tan a = |, ifcotA = T 5 ¥ . 



8 THE EIGHT TRIANGLE. [I, 

§2. TRIGONOMETRIC TABLES. 

The magnitude of an angle is commonly expressed in 
degrees, minutes, and seconds, e.g. 68° 25' 30". A degree is 
the ninetieth part of a right angle ; a minute, the sixtieth 
part of a degree ; a second, the sixtieth part of a minute. 

In the computation of triangles and generally in operations 
that involve angles, the angles themselves play no direct part, 
but the six trigonometric ratios are always used. By methods 
to be explained later, these ratios have been computed for 
different angles and arranged in tables for convenient use. 

In the small tables (pp. ix-xvi) both the ratios themselves, 
the natural functions, and their logarithms, the logarithmic 
functions, are given correct to four figures for angles differing 
by ten minutes, from 0° to 90°. If a logarithm be negative, 
10 is added and the modified logarithm is given. 

The two angles printed on one line are complementary 
angles, and the direct functions of the one are the co- functions 
of the other. Angles less than 45° are found at the left side of 
the page, and the names of their functions at the top ; angles 
greater than 45° are at the right side, and the names of their 
functions at the bottom. 

The functions of an angle given in the tables may be read 
directly from the tables ; but those of an angle not so given 
are found from those of the next less and next greater tabular 
angles, on the principle that small differences of angles and 
the corresponding small differences of functions, are very 
nearly proportional. 

&g.\- sin 25° = .4226, sin 25° 10' =.4252, nearly, [table, 

and sin 25° 5' lies midway between sin 25° and sin 25° 10', 

/.sin 25° 5' = .4230, nearly. 
Stf, v log-tan 25° 20' = 9. 6752, log-tan 25° 30' = 9. 6785, 

.-. log-tan 25° 22' = 9. 6752 + T \ (9. 6785 - 9. 6752) = 9.6759. 

If the functions be given in the table, then the angles may 
be read directly ; but if not so given they maybe found from 
the next less and next greater tabular functions. 



§2.] TRIGONOMETRIC TABLES. 9 

E.g. v cos" 1 .4252 = 04° 50', cos" 1 .4226 = 65°, [table. 

.-. cos" 1 . 4239 = 64° 50' + -H- 10' = 64° 55'. 
So, v log-cotr 1 9.0785 = 04° 30', log-cot" 1 9.6752 = 64° 40', 
.-.log-cot" 1 9. 6759 = 64° 40'-- &. 10' = 64° 38'. 

In the larger tables the decimals are carried to five, or six, 
or seven places, the ratios are given for angles that differ by 
one minute, or by ten seconds, or by one second, and there are 
many labor-saving devices. 

Of these devices, the most common is that of printing the 
differences of consecutive logarithmic sines in a column at the 
right of the column of sines, that of cosines at the right of 
the column of cosines, and that of tangents and cotangents (the 
same differences for both) between the columns of tangents 
and cotangents. These differences are called the tabular dif- 
ferences. 

QUESTIONS. 

From the table of natural functions, find : 

1. sin 20°, 21°, 20° 10', 20° 20', 79° 18', 57° 15'. 

2. cos 20°, 21°, 20° 10', 20° 20', 79° 18', 57° 15'. 

3. tan 35°, 36°, 35° 15', 35° 25', 79° 58', 25° 36'. 

4. cot 35°, 36°, 35° 15', 35° 25', 79° 58', 25° 36'. 
From the table of logarithmic functions, find : 

5. log-sin 20°, 21°, 20° 10', 20° 20', 79° 18', 57° 15'. 

6. log-cos 20°, 21°, 20° 10', 20° 20', 79° 18', 57° 15'. 

7. log-tan 35°, 36°, 35° 15', 35° 25', 79° 58', 25° 30'. 

8. log-cot 35°, 36°, 35° 15', 35° 25', 79° 58', 25° 36'. 
From the table of natural functions, find : 

9. sin" 1 .2588, .2591, .2590, .9279, .9281, .9280. 

10. cos" 1 . 9279, .9281, .9280, .2591, .2588, .2590. 

11. tan- 1 . 5022, .5059, .5035, .9217, .9271, .9250. 

12. cot" 1 . 921 7, .9271, .9250, .5022, .5059, .5035. 
From the table of logarithmic functions, find : 

13. log-sin" 1 8.5809, 8.5842, 8.5821, 9.9997, 9.9847. 

14. log-cos" 1 8.5809, 8.5842, 8.5821, 9.9997, 9.9847. 

15. log-tan" 1 8.5812, 8.5845, 8.5831, 1.4188', 1.3071. 

16. log-cot" 1 8.5812, 8.5845, 8.5831, 1.4188, 1.3071. 



10 



THE RIGHT TRIANGLE. 



P. 



§3. THE SOLUTION OF RIGHT TRIANGLES. 

Three parts, one being a side, are sufficient to determine a 
plane triangle ; and the sohttion of a triangle consists in find- 
ing the three unknown parts from the three that are given. 

In a right triangle if a side and one other part be known, 
the triangle may be solved by forming equations that involve 
the two known parts and one unknown part, and solving these 
equations for the unknown parts. 

In general the work may be checked by forming independent 
equations that involve the three computed parts, and which 
cannot be true unless the work be correct. 

Let abc be a right triangle then, whatever parts be given, 
all the equations needed are found among these : 
a + b = 90°, a' 2 + b 2 = c*, 
sin a = a/c, cos a = b/c, tan a = a/b, 
sin B — b/c, cos B = a/c, tan b — b/a. 






c 


A 




b 










V 




\a 


s 





There are four cases : 

1. Given c, a, the hypotenuse and an acute angle : 
then b = 90° — a, a — c^\n a, 5 = c-cosa. 

Checks: tanB = J/«, b' z — (c + a) (c — a), a 2 = (c + b) (c — b). 

2. Given b, a, a side and an acute angle : 
then B=r90°-A, c = b/eosA, a = b -tan a. 

Checks: cos B = a/c, b 2 = (c + a) (c — a) : a*=(c + b) (c — b). 

3. Given c, b, the hypotenuse and a side : 
then cos A — b/c, b = 90° — a, a = b-t$MiA. 

Checks: cos b = a/c, b 2 ~ (c + a) (c — a), a 2 —(c-\-b) (c — b). 

4. Given a, b, the two sides about the right angle : 
then tan a = a/6, B = 90° — A, c = Z>/cosa. 

Checks: cos b = a/c, b 2 —(c + a)(c — a), a 2 =(c + b) (c — b). 



§3.] THE SOLUTION OF RIGHT TRIANGLES. 11 

E.g. let c = 125, a = 40° ; then b = 90° - 40° = 50° ; 

and, with natural functions, the work may take this form : 



sin 40° = .6428 




cos 40° = .7660 


x 125 




xl25 


a = 80.35. 




I = 95.75. 


check : tan b — b /a, 




(c + b)(c-b) = a 2 


80.35)95.75(1.1917 


C- 


= 125 




b: 


= 95.75 


tan 50° = 1.1918 


C + 1): 


= 220.75 a- 80.35 




C-b: 


= 29.25 x x 80.35 



c 2 - b 2 = 6456.9375 a 2 = 6456.1225 
So, with logarithmic functions, the work may take this form : 

log-sin 40° = 9.808 1 log-cos 40° = 9.8843 

log 125 = 2.0969 + log 125 = 2.0969 + 

log « = 1.9050, « = 80.35 log b = 1.9812, 5 = 95.76 

check: c=125 

b = 95.76 



c + b = 220.76 log, 2.3439 log = 1.9050 

c-b= 29.24 1.4660 x2 

log (c 2 - F) = 3.8099. log a 2 = 3.8100. 

Note. The two solutions do not quite agree, and the checks 
are not perfect ; the defects arise from the use of the small 
tables. More exact results come from larger tables, that give 
the ratios correct to five, six, or seven figures. 

QUESTIONS. 

Solve these right triangles, using natural functions, given : 

1. c, 40 yds.; a, 30°. 2. c, 12,5 ft.;B, 68° 10'. 

3. b, 1S7 metres ; a, 55° 20, 4. a, 7.57 in.; b, 9° 30'. 

5. b, 18.5 ft.; c, 125 ft. 6. c, 37 mi.; a, 25.2 mi. 

7. 0, 59.3 yds.; b, 45.7 yds. 8. 0,4 ft. 6 in.; b, 12ft. 9. in. 
Solve these right triangles, using logarithmic functions, given : 

9. c, 127 ft.; A, 60°. , 10. c, 18.7 yds.; b, 76° 15'. 
11. b, 45.9 yds.; A, 59° 15'. 12, 0, 18.3 chs.; b, 55° 12'. 
13. b, 597 m.; c, 676 m. 14. 0, 1278 yds.; c, 1355 yds. 

15. 0, 27.85 in.; b y 5519 in. 16. 0, 8539 ft.; J, 2815 ft. 



12 



THE RIGHT TRIAXGLE. 



[I, 



§4. ISOSCELES AND OBLIQUE TRIANGLES. 

In an isosceles triangle, the perpendicular from the vertex 
to the base divides the triangle into two equal right triangles ; 
and if two parts of one of these triangles be given, this tri- 
angle may be solved, and so the whole triangle is solved. 

If three parts of an oblique triangle oe given, always includ- 
ing a side, a perpendicular may fall from a vertex to the 
opposite side and so divide the given triangle into two right 
triangles, and by their solution the triangle is solved. 

Let abc be any oblique triangle, a, b, c the sides opposite 
the angles a, b, c ; p the perpendicular ad from a to a ; x, y 
the segments CD, bd, of a. 




The statements below apply directly to the second of the 
three figures ; but with slight modifications suggested by the 
figures themselves, they apply to the other figures as well. 

There are four cases : 

1. Given a, b, c, the three sides : 
then\'p 2 + x 2 = b 2 , p* + y 2 = c\ 

and v x + y — a, 

.-. x — y-{V -&)/a, 

/tX = ^[a + (b 2 -c' i )/a] = (a 2 + b 2 -c' 2 )/2a, 
and y=i [a - (b 2 - c*)/a] = (a 2 - b 2 + c 2 )/2a ; 
and two parts of each right triangle are known. 

2. Given 5, b, c, a side and two angles : 

then, in the right triangle acd, b and care known, and/) and x 

may be computed ; 
and, in the right triangle abd, j# and b are known, and c and y 

may be computed. 
a = x + y, a = 180°-(b + c). 



§4.] ISOSCELES AND OBLIQUE TRIANGLES. 13 

3. Given c, a, b, two sides and the included angle : 

then, in the right triangle abd, c and B are known, and^? and 

y may be computed ; 
and, in the right triangle acd, p is known, x — a — y, and b and 

c may be computed. 
A = 180°-(B + C). 

4. Given &, c, B, two sides and an opposite angle : 

then, in the right triangle abd, c and b are known, and jp and 

y may be computed ; 
and, in the right triangle acd, b and jtf are known, and x and 

c may be computed. 
a = y±x, a = 1'80°-(b + c). 

questions. 
Solve these isosceles triangles, given : 

1. The sides 10 yards, and the base 16 yards. 

2. The vertical angle 90°, and the base 10 yards. 

3. The base 10 yards, and the base angles 70°. 

4. The vertical angle 70°, and a side 12 yards. 

5. The base 18 yards, and a side 12 yards. 

6. If two sides and an angle opposite one of them be given, 
b, c, b, the side c is given in length and position both, a in po- 
sition but not in length, b in length but not in position, and 
b finds its position only as it swings about A as a hinge till its 
lower end rests on the line of the base : if then the angle B be 
acute, and if the swinging side b be shorter than the perpen- 
dicular j9, is a triangle possible ? is there a triangle if b be just 
as long asj9 ? of what kind is it ? is there one triangle or two 
if b be longer than p, but shorter than c ? if b be just as long 
as c ? if b be longer than c ? Draw figures to illustrate. 

So, if b be right or obtuse ? 
Solve these oblique triangles, given : 

7. a, 13 ; b, 15.; c, 17. 8. a, 357 ; b, 537 ; c, 735. 

9. c, 5 j a, 7 ; B, 65°. 10. «, 537 ; 5, 753 ; c, 119° 15 r . 
11. b, 30 ; B, 55°; c, 48° 25'. 12. a, 7.5 ; a, 84° ; b, 42° 37'. 
13. b, 5, 10, 15, 20, 25 in turn ; c, 20 ; B, 30°. 



14 THE RIGHT TRIANGLE. [I, 

§ 5. HEIGHTS AND DISTANCES. 

The plane of the horizon at any point on the earth's surface 
is the plane that is tangent to the surface, i.e. to the surface 
of still water, at that point ; it would therefore be perpen- 
dicular to the radius of the earth, if the earth were a perfect 
sphere. The direction perpendicular to the horizon-plane is 
determined by a plumb line ; it is a vertical line, and any 
plane containing this line is a vertical plane. Any plane 
parallel to the horizon-plane is a horizontal plane, and such a 
plane may be determined by a spirit level. 

An angle lying in a horizontal plane is a horizontal angle, 
and an angle lying in a vertical plane is a vertical angle. The 
vertical angle made with the horizontal plane by the line of 
sight from the observer to any object is its angle of elevation 
if the object be above the observer, and its angle of depression 
if it be below him. 

Ordinary field instruments measure horizontal and vertical 
angles only. By distance is meant the horizontal distance, 
unless otherwise named ; and by height is meant the vertical 
distance of a point above or below the plane of observation. 
A surveyor's chain is four rods long and it is divided into a 
hundred links. Ten square chains make an acre. 

To find the height above its base of a vertical column, ap, 
whose base is accessible. 

1. If the column ap stand on a horizontal plane : 

p 




o A 

From the base a measure any convenient distance AO, and 

the angle aop ; 
and solve the right triangle aop, for ap. 



|5.j HEIGHTS AND DISTANCES. 

2. If the column pq stand on an inclined plane : 



15 




Let p be the top of the column, Q the point at the base of the 
column below p, and A a point below p in the hori- 
zontal plane through the point of observation, o ; 

measure any convenient distance qo along the plane, and the 
angles of elevation or depression aop, aoq ; 

solve the right triangles aop, aoq : then pq = ap± aq. 

To find the distance from the observer, and the height above 
its base, of an inaccessible but visible vertical column. 

Let p be the top of the column, Q the base, b the position of 
the observer, a the point vertically below p in the 
horizontal plane through b ; 




take any other convenient point of observation c, and 
measure the horizontal line bc, the horizontal angles 
abc, acb, and the vertical angles abp, abq ; 

solve the horizontal oblique triangle abc for ab, and the 
vertical right triangles abp, abq for ap, aq : then 

PQ = AP±AQ. 



16 



THE RIGHT TRIANGLE. 



[I, 



If the observer be in the same horizontal plane as the base, 
the line bq coincides with ba, and bap is the only vertical 
triangle to be computed. 

To find the distance apart of tivo objects that are separated 
by an impassable barrier. 

1. If both objects be accessible : 



i^s^>^ 




J^^-T^t 



Let E, f be the two objects, and G the point of .observation ; 
measure the horizontal lines ge, gf and the horizontal angle 
egf, and compute ef. 

2. If both objects be inaccessible : 
Let c, D be the two objects ; measure any convenient line ab 

and the horizontal angles abc, abd, bac, bad ; 
in triangle abd compute bd ; in abc compute bc ; in BCD 
compute CD. 
This is the method of triangulation ; ab is the base line. 

QUESTIONS. 

1. At 120 feet distance, and on a level with the foot of a 
steeple, the angle of elevation of the top is 62° 27' : find the 
height. [230.03 feet. 

2. From the top of a rock 326 feet above the sea, the angle 
of depression of a ship^s hull is 25° 42' : find the distance of 
the ship. [677.38 feet. 

3. A ladder 29J feet long standing in the street just reaches 
a window 25 feet high on one side of the street, and 23 feet 
high on the other side : how wide is the street ? [34.13 feet. 



§5.] HEIGHTS A^D DISTANCES. 17 

4. From the top of a hill I observe two successive mile- 
stones in the plain below, and in a straight line before me, 
and find their angles of depression to be 5° 30', 14° 20' : what 
is the height of the hill ? [815.85 feet. 

5. Two observers on the same side of a balloon, and in the 
same vertical plane with it, are a mile apart, and find the 
angles of elevation to be 17° and 68° 25 ' respectively : what 
is its height ? [1836 feet. 

6. From the top of a mountain 1J miles high, the clip of 
the sea-horizon (angle of depression of sky-and-water line) is 
1° 34' 40" : find the earth's diameter, and the distance of the 
sea-horizon. 

7. What is the distance and the dip of the sea-horizon from 
the top of a mountain 2f miles high, the earth's mean radius 
being 3956 miles ? [2° 8' 8". 

8. If the dip of the sea-horizon be 1°, find the height of 
the mountain, and the distance of the sea-horizon. 

9. How far should a coin an inch in diameter be held from 
the eye to subtend an angle of 1° ? 

10. Given the earth's equatorial radius, 3962.76 miles, and 
the angle this radius subtends at the sun, 8". 81 : find the dis- 
tance of the earth from the sun. [92 780 000 miles nearly. 

11. Find the distance across a river, if the base ab be 475 
feet ; the angle A, 90° ; the angle b, 57° 13' 20". [737.68 feet. 

12. Given ca, 131 feet 5 inches ; bc, 109 feet 3 inches ; 
the angle c, 98° 34' : what is the distance ab ? [183 feet. 

13. Two ships lying half a mile apart, each observes the 
angle subtended by the other ship and a fort ; the angles are 
found to be 56° 19' and 63° 14' : find the distances of the 
ships from the fort. [2525, 2710 feet. 

14. Given the base ab, 1314- yards ; the angle bad, 50° ; the 
angle bac, 85° 15' ; the angle dbc, 38° 43'; the angle dba, 94° 
13' : what is the distance cd ? Check the work by making 
two distinct computations from the data. [129.99 yards. 



18 



THE RIGHT TRIANGLE. 



[I> 



§6. COMPASS SURVEYING. 

In compass surveying, the bearing of a point is the hori- 
zontal angle which the line of sight from the observer to the 
point makes with the north-and-south line through the point 
of observation. This angle is found by aid of the compass. 

The latitude of a point is its distance north or south of a 
given point. The latitude of a line is the length of its pro- 
jection on a north-and-south line ; and its departure is the 
length of its projection on an east-and-west line. 
E.g. in the figure below, representing a field, the starting 
point is A, the bearings of the lines ab, bc- • •, taken 
in order, are : s. 70° 20' e. (70° 20' east from south^ 
s. 10° 15' e., k. 55°35'e., H". 18°45'w„ s. 40°55'w., 
s. 37° 15' w.; 
and the lengths of these lines, in chains, are : 6.37, 4.28, 12.36, 
14.96, 11.15, 8.00. 




Through all the points A, B • ■ * , are drawn north-and- 
south lines, marked on the figure with arrows, and east-and- 
west lines perpendicular to them. The north-and-south line 
through the starting point A is distinguished as the meridian. 

The latitude of ab is the length of ab', the projection of 
ab on the meridian, and it is computed by multiplying 6.37, 
the length of ab, by the cosine of 70° 20', the bearing of ab. 



§c] 



COMPASS SURVEYING. 



19 



So, the departure of ab is the length of b'b, i.e. the prod- 
uct of 6.37 by the sine of 70° 20'. 

The latitude of the line bc is the length of bc", i.e. the 
product of 4.28 by the cosine of 10° 15', and the departure of 
bc is the length of c"c, i.e. the product of 4.28 by the sine 
of 10° 15'; and so for the latitudes and departures of the 
other lines, as shown in the table below. 

The meridian distance of a point is the distance of the 
point east or west from the meridian, and the double meridian 
distance of a line is the sum of the meridian distances of its 
ends. 

E.g. the meridian distance of the point B is b'b, and that of 
c is c'c, which is equal to b'b + c"c. 

So, the double meridian distance of the line ab is + b'b, 
and that of bc is b'b + c'c. 

When a surveyor has run round a field, e.g. that which is 
described above, and has found and set down the lengths and 
bearings of the sides, he has next to compute the latitudes 
and departures of the sides, the meridian distances of the cor- 
ners, and the double meridian distances of the sides as shown 
above. He is then ready to compute the areas of certain 
trapezoids and right triangles, and finally the area of the 
field ; and he takes care to set down his work in such form 
that it can be easily understood and reviewed, generally in 
the form of a table as below. 





BEARING. 


DIS- 
TANCE . 


DEP. 


M.D. 


D.M.D. 


LAT. 


DOUBLE AREA. 

+ — 


AB 


S. 70° 20' E. 


6.37 


5.998 


5.998 


5.998 


-2.144 




-12.860 


BC 


s. 10° 15' e. 


4.28 


.761 


6.759 


12.757 


-4.212 




-53.732 


CD 


x. 55° 35' e. 


12.36 


10.196 


16.955 


23.714 


6.985 


165.642 




DE 


N. 18° 45' w. 


14.96 


-4.809 


12^0 
4^3 


29.101 


14.166 


412.245 




EF 


s. 40° 55' w. 


11.15 


-7.303 


16.989 


-8.426 




-143.149 


FA 


s. 37° 15' w. 


8.00 


-4.843 


0. 


4.843 


-6.369 




-30.845 










+577.887 -240.586 
















-240.586 





337.301/2=168.651 square chains=16.865 acres. 337.301 



20 



THE RIGHT TRIANGLE. 



EX 



In this figure there are two right triangles ab'b, ff'a and 
four trapezoids bb'c'c, cc'd'd, dd'e'e, ee'f'f, so related that 
the area of the polygon abcdef is the excess of the sum of 
the two trapezoids cc'd'd/ dd'e'e over the sum of the two tri- 
angles and the other two trapezoids. 

i.e. abcdef = — ab'b — bb'c'c -h cc'd'd + dd'e'e — ee'f'f — ff'a 
= | [-ab'-b'b-b'c'- (b'b + c'c) + cV- (c'c + d'd) 
+ d'e' • (d'd + e'e) — e'f' • (e'e + f'f) — f'a • f'f] 

and it remains only to compute the lines ab', b'b* • •, and to 
add, subtract, and multiply as shown below. 



E'|_ 




IE 




1 
1 

1 

J v 

F[_ Fj/ 

i A 




/ l\ 

i \ 
1 \ 
l \ 
1 \ 
\ 
\ 

If" \ 




/ 
» / 
3 / 

Dl / 


ID 




D 


Bl B\ 














c'i r 7 


C 







In detail the work may take this form : 

1. To compute the latitudes and departures of the sides: 



s. 70° 20' e. 


s. 10° 15' e. 


is. 55° 35' e. 


cosine sine 


cosine sine 


cosine sine 


.3365 .9417 


.9840 .1779 


.5652 .8249 


6.37 ab 6.37 


4.28 bc 4.28 


12,36 CD 12.36 


-2.144 5.998 


-4.212 .761 


6.986 10.196 


N. 18° 45' w. 


s. 40° f 5' w. 


s. 37° 15' w. 


cosine sine 


cosine sine 


cosine sine 


.9469| .3214| 


.75561 .6550 


.7960 .6053 


14.96 de 14.96 


11.15 EF 11.15 


8.00 fa 8.00 


14.166 "4.809 


-8.426 -7.303 


-6.369 "4.843 



§0.] COMPASS SURVEYING. 21 

North latitudes, northings, are called positive ; south lati- 
tudes, southings, negative. East departures, eastings, are 
called positive ; west departures, westings, negative. 

2. To compute the meridian distances : 

B C D E 

5.998 5.998 6.759 16.955 

+ .761 + 10.196 - 4.809 
6.759 16.955 12.146 4.843 

3. To compute the double meridian distances : 



F 


A 


12.146 


4.843 


- 7.303 


-4.843 



AB 


BC 


CD 


DE 


EF 


FA 


5.998 


5.998 


6.759 


16.955 


12.146 


4.843 




+ 6.759 


+ 16.955 


+ 12.146 


+ 4.843 





12.757 23.714 29.101 16.989 

4. To compute the double areas : 

abb' bb'c'c cc'd'd dd'e'e ee'f'f ff'a 

5.998 12.757 23.714 29.101 16.989 4.843 

x -2.144 x -4.212 x +6.985 x +14.166 x ~8.426 x "6.369 

-12.860 53.732 +165.642 +412.245 "143.149 "30.845 

questions. 

1. A surveyor, starting from A, runs n - . 22° 37' e. 3.37 
chains to b; thence :n t . 80° 24' e. 3.81 chains to c ; thence s. 
41° 12' E. 5.29 chains to d ; thence s. 62° 45' w. 6.22^ chains 
to e : find the latitude and meridian distance of b, c, d, e 
from A ; find the bearing and distance of A from e ; find the 
area of the field abcde. 

2. Starting at A and chaining along the surface of the 
ground, a surveyor runs isr. 81° 10' e. 48 chains to b, at an 
elevation of 4° 15' ; thence n. 26° 25' w. 126 chains to c, at 
an elevation of 3° 40' ; thence s. 73° 50' w. 45 chains to d, 
at an elevation of 2° 40' ; thence s. 60° e. 85 chains to E, at a 
depression of 4° 15' : find the horizontal distances ab, bc, cd, 
de, and the heights of b, c, d, e abave A ; find the bearing, 
distance, and angle of depression, of A from e ; find the area 
of the field abcde. 



22 GENERAL PROPERTIES OF PLANE ANGLES. [II, 

II. GENERAL PROPERTIES OF PLANE ANGLES. 



Hitherto the lengths of the sides of a triangle and the 
magnitudes of the angles have been mainly considered, and 
little attention has been paid to their directions ; but greater 
generality, as well as greater definiteness, is given to the 
definitions and theorems of trigonometry if the lines and 
angles be thought of as directed as well as measured. 

Nor is this a new thing : in geography and navigation 
longitudes are distinguished by the words east and west, and 
latitudes by north and south ; a surveyor speaks of his 
northings and southings and of his eastings and ivestings, 
and he writes down the bearings of his lines with the sig- 
nificant letters N, s, E, w ; in physics the directions and inten- 
sities of forces are represented by the directions and lengths 
of lines. 

Even the language is not new : the mathematician merely 
makes use of the familiar algebraic words positive and nega- 
tive as more convenient to him than the commoner words 
north, south, east, west, up, down, right, left, forward, 
backward. 

§1. DIRECTED LINES. 

Hereafter every straight line will be regarded as having not 
only position but direction also, meaning thereby that a point 
moving along the line one way will be regarded as moving 
forward, and a point moving along the line the other way as 
moving backward. The direction of the line is assumed to 
be that of forward motion. 

If a line represent a force or an actual motion, like that of 
the winds and the tides, it has a natural direction ; otherwise 
its direction may be assumed at will. 

E.g. with a double-track east-and-west railway, the south 
track may be used habitually by east-bound trains, and the 
north track by west-bound trains. On the south track a 
train moves forward when going east, and it goes west only 



§ 1.] DIRECTED LIKES. 23 

when backing. On the north track forward motion is west- 
ward motion. The two tracks may be regarded as two par- 
allel lines lying close together and having opposite directions. 

A segment of a line is a limited portion of the line that 
reaches from one point, the initial point of the segment, to 
another point, the terminal point. A segment is a positive 
segment if it reach forward, in the direction of the line, and 
a negative segment if it reach backward. 

It is convenient also to speak of the positive and negative 
ends of a line, meaning by the positive end that end which 
is reached by going forward along the line, from any start- 
ing point upon it, and by the negative end that end which 
is reached by going backward. 

E.g. if a north-and-south line be directed from south to 
north, then the north end is the positive end and the south 
end is the negative end of the line; segments of this line 
reaching northward are positive segments and segments reach- 
ing southward are negative segments. 

The direction of a line is indicated by an arrow, or by 
naming two of its points, the direction being from the point 
first named towards the other. The direction of a segment 
is shown by the order of the letters at its extremities, the 
initial point being named first and the terminal point last. 

E.g. the indefinite line op has its positive direction from 
o to p, and the segment ab of the line op is the segment 
that reaches from the point A to the point B. 

■5 "5 1 !p "i- K 

If two segments, not necessarily upon the same line, have 
the same length and be both positive or both negative, they 
are equal segments ; if they have the same length, and be one 
positive and the other negative, they are opposite segments. 

ADDITION OF SEGMENTS OF A STRAIGHT LINE. 

Two or more segments of a straight line are added by 
placing the initial point of the second segment upon the 
terminal of the first, the initial point of the third segment 



24: GENERAL PROPERTIES OF PLANE ANGLES. [II, 

upon the terminal of the second, and so on ; and the sum of 
all the segments so added is the segment that reaches from 
the first initial to the last terminal point. When a positive 
segment is added, the terminal point slides forward ; when 
a negative segment is added, it slides backward. 
E.g. in the figures below, 

AB + BA = 0, AB-f-BC=AC, AB + BC + CA = 0, 
AB + BC + CD = AD, AB + BC + CD + DA = 0. 



ABC D 

l 1 1 >. 

BCD A 

I 1 1 ^ 

CAB D 

This addition is analogous to the addition of like numbers, 
positive and negative, in algebra. 

One segment is subtracted from another by adding the op- 
posite of the subtrahend to the minuend, or by placing the 
initial point of the subtrahend upon that of the minuend ; 
the remainder is then the segment that reaches from the ter- 
minal point of the subtrahend to that of the minuend. 

QUESTIONS. 

1. If from a given starting point one man w r alk east and 
another west, each a hundred yards, how far apart are the 
two men ? how far, and in what direction, is the first man 
from the second ? the second man from the first ? 

2. If the river run five miles an hour, how fast does a boat 
go, with the current, if the crew can row four miles an hour 
in still water ? against the current ? 

3. If longitudes alone be under consideration, and west 
longitudes be marked -f , how may east longitudes be marked ? 
how may north and south latitudes be then distinguished ? 

4. If a traveller go east 50 miles, then west 30 miles, then 
west 60 miles, then east 20 miles, how far has he gone ? and 
how far, and in what direction, is he from the starting point ? 



§2.] DIRECTED PLANES AND ANGLES. 25 

§2. DIRECTED PLANES AND ANGLES. 

Hereafter every plane will be regarded as having direction, 
meaning thereby that a line swinging about a point in the 
plane one way will be regarded as swinging forward, and a line 
swinging the other way as swinging backward. The direc- 
tion of the plane is that of the forward motion of the line. 

If the swinging line has a natural motion like that of the 
hands of a clock, or a spoke of the fly-wheel of an engine, or 
an equatorial radius of the earth, then the direction of the 
plane is determined by this motion ; otherwise its direction 
may be assumed at will. 

This swinging motion, as viewed from one side of the 
plane, is clockivise, i.e. left-over-right, and counter -clock wise, 
i.e. right-over-left, as viewed from the other side. 

E.g. the apparent daily motion of the sun, as seen by an ob- 
server in the northern hemisphere, is clockwise, 

and as seen by one in the southern hemisphere it is counter- 
clockwise ; 

but to both of them it is the same east-to-west motion, and the 
plane of the sun's apparent path is an east-to-west plane. 

So, the real motion of an equatorial radius of the earth is 
counter-clockwise if viewed from a point in the 
northern hemisphere, 

and clockwise if from a point in the southern hemisphere ; 

but it is the same west-to-east motion, and the plane of the 
equator is a west-to-east plane, whose direction is 
opposite to that of the sun's apparent path. 
An observer to whom forward motion appears counter-clock- 
wise is in front of the plane, and looks at its face ; one to 

whom forward motion appears clockwise is hack of the plane. 

E.g. the plane of the equator faces northward, and points in 
the northern hemisphere are in front of it ; 

but the plane of the sun's apparent path faces southward. 
In plane trigonometry the reader always looks at the face 

of his plane, and to him, therefore, forward motion is always 

counter-clockwise motion. 



2G GENERAL PROPERTIES OF PLANE ANGLES. [II, 

DIRECTED ANGLES. 

& plane angle has been variously defined as " the opening 
between two lines," as "the inclination of one line to another/' 
as "the difference of direction of two lines," and as "the por- 
tion of the plane between the two lines." The words "incli- 
nation " and "difference of direction" appear to define the 
magnitude of the angle rather than the angle itself ; but 
whichever of these definitions be used, it is manifest that an 
angle may be generated by swinging a line, in the plane of 
the angle, about the vertex, from one of its bounding lines to 
the other. The first position of the swinging line is the ini- 
tial line, and the last position is the terminal line, of the angle. 
E.g. the minute-hand of a clock generates a right angle every 
fifteen minutes, and four right angles in an hour. 

If the generating line swing forward, in the direction of 
the plane, it generates a positive angle ; if it swing backward, 
it generates a negative angle. 

Since, in plane trigonometry, the reader always looks at the 
face of his plane, it follows that positive angles are counter- 
clockwise angles, and negative angles are clockwise angles. 

Tlie angle of two lines is the smaller of the two angles 
which lie between their positive ends and reaches from the posi- 
tive end of the line first named to the positive end of the other. 
E.g. if the two lines a'a, b'b cross at o, the angle of the two 
lines a'a, b'b is aob, and the angle of the two lines b'b, 
a'a is BOA. 



B B 



A A 



The two bounding lines may be designated by single letters, 
the initial line being named first. 

E.g. if 7, m stand for the two lines a'a, b'b, then Im stands 
for the angle aob and ml for the angle boa. 





:2-] 



DIRECTED PLANES AND ANGLES. 



27 



NORMALS. 



One line is normal to another if the first line make a posi- 
tive right angle with the other. 
E.g. in the figures below, ob is normal to oa, but not oa to ob. 




,A 




A 




k 




/ 

1 




B 









B A 




EQUAL AND CONGRUENT ANGLES. 

If two angles differ by one or more complete revolutions, 
they are congruent ; if, when placed one on the other, their 
initial lines coincide and their terminal lines coincide, they 
are equal or congruent. 




E.g. in the figures above all the angles aob, whether positive 

or negative, are congruent, 
and the angles aob, a'ob' are equal, but not aob, b'oa'. 



23 GENERAL PROPERTIES OF PLANE ANGLES. [II, 

The smallest angle, positive or negative, of a series of con- 
gruent angles is the primary angle ; and the primary angle 
is always meant if no other be indicated. It is always smaller 
than two right angles. 

QUESTIONS. 

1. If a surveyor by mistake write n. 30° e. 12 chains, 
instead of N. 30° w. 12 chains, what is his error ? and what 
is the effect, in his map, on the position of every subsequent 
line and point ? 

2. If the line a be normal to the line b, what angle does 
b make with a ? 

3. Through what angle has the hour-hand of a clock swept 
from 12 midnight to 12 noon ? the minute-hand ? the second- 
hand ? 

4. If the moon revolve about the earth once in four weeks, 
what is its angular motion in a year ? in a day ? 

5. How great is the angular motion of the earth upon its 
own axis in a day ? in an hour ? in a year ? 

So, how T great is its angular motion in its orbit about the 
sun in a year ? in a day ? in a century ? 

C. What is the angle between a north wind and a north- 
east w r ind ? a north wind and a southwest wind ? 

7. If the current carry a chip due south, and the wind 
carry a feather due east, what is the angle between the arrows 
that show the directions of the motions of the chip and the 
feather ? 

8. If two forces act upon a body, the one vertical and the 
other horizontal, what is the angle between them ? its sign ? 

9. If three equal forces acting upon a body be parallel to 
the three sides of an equilateral triangle, what are the 
angles between them ? Discuss the eight possible cases. 

10. If the two hands of a clock start together at noon, 
what is the angle between them at one o'clock ? at two ? at 
three ? at six ? at nine ? at twelve ? 



*•] 



DIRECTED PLANES AND ANGLES. 



29 



ADDITION OF CO-PLANAR ANGLES. 

Two or more co-planar angles are added by placing the ini- 
tial line of the second angle upon the terminal of .the first, the 
initial line of the third angle upon the terminal of the second, 
and so on ; and the sum of all the angles so added is one of 
the congruent angles reaching from the first initial to the last 
terminal line. This definition applies whether the vertices 
of the angles be at the same point or at different points. 

AVhen a positive angle is added, the terminal line swings 
forward ; when a negative angle is added, it swings backward. 
E.g. in the figures below, 




ab + ba = (or one of the congruents of 0), ab + bc = ac, 
ab + bc + ca = 0, ab + be + cd = ad, ab + be 4- ecl 4- da — 0. 

One plane angle is subtracted from another by adding 
the opposite of the first angle to the other, or by placing the 
initial line of the first angle upon that of the second ; the 
remainder is then the angle that reaches from the terminal 
line of the first angle to that of the other. 

If the sum of two angles be a positive right angle, either 
angle is the complement of the other ; and if their sum be two 
right angles, either angle is the supplement of the other. 



30 



GENERAL PROPERTIES OF PLANE ANGLES. [II, TH. 



QUESTIONS. 

1. Show what angles must be added to the angles below to 
make the sums positive right angles, and so construct their 
complements. 



->- V-J > -< *_o_J >. {. \ > I rA 



- m 



_*__^, \ v > 



*- ( r iL -^ ' 



\ 4* — >- 



2. Show what angles must be added to these angles to make 
the sums two positive right angles, and so construct their 
supplements. 

3. Show that the angle of two lines equals the angle of 
any normals to them. 

4. If a surveyor, in running round a field, turn at the cor- 
ners always to the left, what is the sum of the exterior angles 
of the field ? if he turn always to the right ? if he turn some- 
times to the right and sometimes to the left ? 




IS, 





/ 


1 > , 


('/ 








\ 


' / 




5. If the wind shift from north to northeast, and then 
from northeast to southeast, through what angle has it 
shifted? 



1, §3.] PROJECTIONS. 31 

§3. PROJECTIONS. 

The orthogonal projection of a point upon a line is the foot 
of the perpendicular from the point to the line ; and, in this 
book, by projection is always meant orthogonal projection. 
The line on which the projection is made is the line of pro- 
jection, and the perpendicular is the projecting line. 

The projection of a segment of one directed line upon 
another directed line is the segment of the second line that 
reaches from the projection of the initial point of the given 
segment to the projection of its terminal point. The projec- 
tion is positive if it reach forward in the direction of the line 
of projection, and negative if it reach backward ; its sign 
may be like or unlike that of the projected segment. 
E.g. the shadow of a post on a plane perpendicular to the 
sun's rays is an orthogonal projection of the post. 

Projections upon the same line are like projections. 

Theor. 1. Jf segments of one directed line he projected upon 
another such line, the ratios of the projections to the segments 
are equal. 

Let l> m, be any two directed lines ; take ab, cd, ef- • • 
segments of m, and let a'b', c'd', e'f'- • • be their 
projections on I ; 

then will a'b'/ab=:c'd'/cd = e'f'/ef« • • 




For v the projecting lines aa', bb'- • • are all parallel, 
and contrary segments of the same line have contrary pro- 
jections on another line, 
.*. the segments and their projections are proportional ; 
i.e. a'b'/ab = c'd'/cd = e'f'/ef = oa'/oa = of'/of • • • , both 
in magnitude and sign. q.e.d. 



32 



GENERAL PROPERTIES OF PL AXE ANGLES. [II, TH. 



In the first figure the segments ab, ef are positive, and so 
are their projections a'b', e'f', but CD, c'd' are negative, and 
all the ratios are positive. In the other figure ab, c'd', ef 
are positive, but a'b', cd, e'f' are negative, and all the ratios' 
are negative ; i.e. the ratios are positive if the primary angle 




of the two lines be acute, positive or negative ; they are nega- 
tive if the primary angle be obtuse. 

Cor. 1. Equal segments of one line have equal projections 
on another line, and opposite segments have opposite projections. 

Cor. 2. If on each of tivo directed lines equal segments of 
the other line be projected, the projections are equal. 




E.g. let I, m be any two lines, ab a segment of I, and cd, 
ef segments of m equal to ab, 

let a'b' be the projection of ab on m and c'd', e'f' the projec- 
tions of cd, ef on I, 
then a'b', c'd', e'f' are equal in magnitude and sign. 

In the first figure the segments and their projections are all 
positive ; in the other figure the segments are negative, and 
their projections are positive. 

Cor. 3. If there be two equal angles, and if equal segments 
of the bounding lines be projected, each upon the other bound- 
ing line of its angle, these projections are equal. 
For the two figures may be placed one upon the other, and 
then cor. 3 becomes a case of cor. 2. 



1, §3.] PROJECTIONS. 33 

QUESTIONS. 

1. A line is 5 feet long and its projection on another line, 
a, is 4 feet long : how long is its projection on a normal to a ? 

Can the signs of the projections be found from the data ? 

2. If a, b be two directed lines at right angles to each 
other, how long is that segment whose projections on a, i are 
5 feet and 12 feet ? — 5 feet and —12 feet ? 

Can the sign of the segment be found from the data ? 

3. Construct lines so that segments of one being projected 
on the other, the ratios of the projections to the segments 
shall be 1, f, f, h h h *>> ~h "h ~h ~ 1 > in turn - 

4. A pole ten feet long points northward and makes an 
angle of 45° with the level ground : how long is its shadow, 
if the sun be directly overhead ? 

So, how long is its shadow on a north-and-south wall, at 
sunrise, if the sun rise due east ? 

Of these two shadows, which is the longer ? 

So, which is the longer if the inclination be 60° ? 

From what point of view would the pole appear to be ver- 
tical ? from what point horizontal ? 

5. Describe an isosceles triangle by walking due east 100 
yards, then northwest 70.7 yards, then southwest 70.7 yards, 
thus giving direction to the sides. 

Project the two sides of this triangle upon the base : what 
relation have these two projections ? 

So, project these two sides upon the bisector of the vertical 
angle : what relation have the two projections now ? 

6. In an equilateral triangle, whose sides are directed by 
walking about it and turning to the left at the vertices, how 
do the projections of the sides upon the base compare in 
length ? in sign ? 

So, the projections upon a normal to the base ? 

7. Can a segment of a line be so projected upon another 
line, that the projection is longer than the segment itself ? 

8. Taking note of signs, what is the range of magnitude 
for the ratio projection/segment ? segment/projection ? 

3 



34 



GENERAL PROPERTIES OF PLANE ANGLES. 



EH 



§ 4. TRIGONOMETRIC RATIOS. 

If a segment of the terminal line of an angle be projected 
upon the initial line and upon a normal to the initial line, 
the first projection may be called the major projection of the 
segment, the other the minor projection, and the ratios of 
these two projections to the segment and to each other are 
named as below : 

minor projection/segment, the sine of the angle, 
major projection/segment, the cosine, 
minor projection/major projection, the tangent, 
major projection/minor projection, the cotangent, 
segment/major projection, the secant, 
segment/minor projection, the cosecant. 

These definitions apply to all angles whatever their magni- 
tudes or signs, and they include as a special case the defini- 
tions given on page 2. 



Y> 


< 


Yrf 


y 




1 ^- 




^P 




; 


,Y 




\r 




y 




i 
i 
i 


V,- 


I— V 



O X 



Y 


k o 


X 






^v>- 


^\r* 


X 


y 












P 



E.g. in the figures above, let xop be any angle a ; let oy be 
normal to the initial line ox, r any segment of the ter- 
minal line op, x, y the major and minor projections of r ; 

then sin a = y/r, cos a = x/r, tan a = y/x, 
esc a = r/y, sec a = r/x, cot a — x/y. 



§4.] TRIGONOMETRIC RATIOS. 35 

The segment r may be taken positive or negative ; for if 
the segment be reversed both projections are reversed, and 
the ratios are unchanged. 

Two other functions in common use are the versed sine 
and coversed sine ; they are defined by the equations 
vers a — 1 — cos a, covers a = l — sin a. 

QUESTIONS. 

1. How do the major and minor projections of the seg- 
ment of a line compare in length with the segment itself ? 
how with each other ? 

2. Can the sine of an angle be larger than 1 ? as large as 
1 ? smaller than 1 ? can the sine be naught ? the cosine? 

3. Can the tangent of an angle be naught ? can it be 
smaller than 1 ? as large as 1 ? larger than 1 ? how large may 
the tangent be ? the cotangent ? the secant ? the cosecant ? 

4. What relations as to sign have a segment and its pro- 
jections ? Draw figures in which : 

all three are positive ; all three negative ; 

the segment and major projection are positive and the 

minor projection negative ; 
the segment is negative and both projections positive. 

5. If two lines be parallel and like directed, what is their 
angle ? How long is the major projection of a segment of 
one of these lines as to the other ? the minor projection ? 

What are the ratios of this angle ? 
So, if two parallel lines have opposite directions ? 
So, if the terminal line be normal to the initial line ? 
So, if the initial line be normal to the terminal line ? 

6. Construct the angles ^R, — ^R, fR, — fR, f R, — |r 
and find their ratios. [R = a right angle. 

Which of these angles have the same sines ? the same 
cosines ? the same tangents? the same secants ? 

So, for the angles ^-R, — Jr, f r, — f R, | R, - f R, f R, — f R. 

7. Construct sin -1 ^, — f, f, 1, 0; cos -1 f, ± \-, — 1 ; 

tan -1 J, |, 0,-1, oo ; cot -1 \, - f , ± 1 . 



36 



GENERAL PROPERTIES OP PLANE ANGLES. [II, TH. 



ANGLES IN THE FOUR QUARTERS. 

If there be two lines such that the second line is normal 
to the first, the plane of these lines is divided into four quar- 
ters. The first quarter lies between the positive ends of the 
two lines, the second quarter between the positive end of the 
normal and the negative end of the first line, the third quar- 
ter between their negative ends, the fourth quarter between 
the negative end of the normal and the positive end of the 
first line. 

An angle is an angle in the first quarter, in the second quar- 
ter, in the third quarter, or in the fourth quarter, according 
as its terminal line lies in the first, second, third, or fourth 
quarter, counting from the initial line. 

It is therefore an angle in the first quarter if its primary 
congruent angle be a positive acute angle ; in the second 
quarter, if a positive obtuse angle ; in the third quarter, if a 
negative obtuse angle ; in the fourth quarter, if a negative 
acute angle. 




E.g. of the figures above, the first angle and the eighth are 

angles in the first quarter, 
the second and seventh are angles in the second quarter, 
the third and sixth are angles in the third quarter, 
the fourth and fifth are angles in the fourth quarter. 



2, § 4 -] 



TRIGONOMETRIC RATIOS. 



37 



POSITIVE AND NEGATIVE RATIOS. 

Theor. 2. The trigonometric ratios of an angle in the first 
quarter are all positive. 

The sine and cosecant of an angle in the second quarter are 
positive; the cosine, secant, tangent, cotangent are negative. 

The tangent and cotangent of an angle in the third quarter 
are positive ; the sine, cosecant, cosine, secant are negative. 

The cosine and secant of an angle in the fourth quarter are 
positive; the sine, cosecant, tangent, cotangent are negative. 

For if r be taken positive, and x, y be the major and minor 
projections of r ; 



Y, 


< 


P^ 


y 


^1 1 


1 >, 





o X 



Y) 


b 


X 








1 


X 


y 




^\r 










P 



then v in the first quarter r, x, y are all positive, 

.-.the ratios y/r, r/y, x/r, r/x, y/x, x/y, are all 
positive ; 
and v in the second quarter r, y are positive, and x negative, 

.*. the ratios y/r, r/y are positive, the rest negative ; 
and v in the third quarter r is positive, and x, y negative, 

.'. the ratios y/x, x/y are positive, the rest negative ; 
and v in the fourth quaiter r, x are positive, and y negative, 

.*. the ratios x/r, r/x are positive, the rest negative. 



38 



GENERAL PROPERTIES OF PLANE ANGLES. [II, THS. 



QUESTIONS. 

Show what quarters these angles lie in, and what signs their 
ratios have : [r = a right angle. 

1. £r, -Je, |r,-§r, §r,-|r, |r,_Jr... 

2. JR, - 1R, I R, - f R, JR, - J Ri Y R , - Y^ . . . 

3. |R,-fR, |R,-fR, |B f -|B, Y^— V^... 

4. 100°, 200°, 300°, 400°, 500°, 600°, 700°, 800°, 900°. 

5. "165°, -365°, -565°, "765°, "965°, -1165°, "13G5°. 
Construct the angles below, and find the values of : 

6. sin 225°, 585°, 810°, 960°, "225°, "585°, "960°. 

7. cos 315°, 675°, 960°, 1110°, 

8. tan 495°, 945°, 1110°, 1260°, 

9. cot 675°, 1035°, 1260°, 1410°, 
10. sec 855°, 1215°, 1410°, 1560°, 



-315°, "675°, -1110°. 

-495°, "915°, -1260°. 

-675°, -1035°, -1410°. 

-855°, -1215°, "1560°. 



11. esc 1035°, 1395°, 1560°, 1710°, "1035°, "1395°, "1710°. 

§ 5. RELATIONS OF RATIOS OF A SINGLE ANGLE. 

Theor. 3. The square of a segment is the sum of the 
squares of its projections on a line and a normal to the line. 




For these projections are equal to the sides of a right tri- 
angle whose hypotenuse is the given segment. 



3, 4, §5.] RELATIONS OF RATIOS OF A SINGLE ANGLE. 



39 



Theor. 4. If ale any pla?ie angle, then: 

sin a -esc a — \, cos a*sec a = l, tan a-cot a = l ; 

tan a — sin a /cos a, cot a — cos a /sin a ; 

sin*a + cos'a = l, sec*a=l + ta?i 2 a, esc 2 a = 1+ cot 2 a. 

For let r stand for any segment of the terminal line of the 

angle, and x, y for its major and minor projections ; 
thenv sin a = y/r, cos a — x/r, tan a = y/x, 

esc a — r/y, $eca = r/x, cota = x/y, [df. 

/.sin a -esc a — 1, cos # • sec c* = 1, tan a>cot a — \ ; 

sin a'/cos a — y/r : x/r = y/x = tan a, 

cos cr/sin a — x/r : y/r = x/y — cot cr. 

a; 2 + y 2 = r\ [theor. 3. 

*/** + tfh* = h x + # 2 A 2 = r2 A 2 > ^/*/ 2 + i = r 7/> 

cos*a-\-sm*a = l, l + tan 2 ^ = sec 2 ^ cot 2 or + 1 = csc 2 cr. 
Cor. If a be any plane angle, then: 



and 



So, 



t.e 



sin a— 


cos a — 


fom a=r 


cot a= 


SCCa=r 


CSC<2=: 


sin a 

V(l— cos* a) 
tan a 


\/(l—sin i a) 

cos a 

1 


SMi a: 


\/(l — sin' 2 a) 


1 


1 


j/(l-5m 2 a) 
V(l — cos 2 a) 


sin a 
cos a 

\/(l — cos- a) 

1 

tana 

cot a 

1 


y(l— sin* a) 

1 

cos a: 

y(to 2 a: + 1) 

|/(C0^ 2 « 4-1) 


sma: 
1 


cos a 

1 

cot a 

|/(sec 2 a — 1) 

1 
^/ (csc 1 a— I) 


\/(l — cos 2 a) 
\f (tan^a + 1) 


y{tan*a + l) 
1 


y '{tail 1 a + 1) 
cot a 


tana 

*/(coPa + V) 
sec a 


\/{cot' 2 a + l) 
^(setfa— 1) 


\'{coPa + 1) 

1 

sec a 

^(cstfa— 1) 


cot a 
sec a 
esc a 


sec a 
1 


^/{sec z a — 1) 
y(csc 2 a— 1) 


^/(sec v a—l) 


esc a 


esc a 


j^{csc' 2 a— 1) 


esc a 



The proof of these equations is left as an exercise for the 
pupil, but certain relations may be noted : 
The values of the cosecant set down in the sixth column are 

reciprocals of those of the sine in the first, 
those of the secant in the fifth column of those of the cosine 
in the second, 



40 GENERAL PROPERTIES OF PLANE ANGLES. [II, TH. 

and those of the cotangent in the fourth column of those of 
the tangent in the third. 

The values of the tangent and cotangent set down in the third 
and fourth columns are quotients of the values of the 
sine and cosine in the first and second columns. 

QUESTIONS. 

1. For a given value of the sine, how many values has the 
cosecant ? the cosine ? the secant ? the tangent ? the cotan- 
gent ? 

What signs have the radicals in each of the four quarters ? 

2. For a given value of the cosecant, how many values has 
each of the other five ratios ? 

3. So, for a given value of the cosine ? of the secant ? of 
the tangent ? of the cotangent ? 

4. Construct the two angles whose sines are + f, and show 
that the two cosines are +f and — f. 

5. Construct the two angles whose cosines are — |, and 
show that the two sines are +| and — f. 

6. Construct the two angles whose tangents are +f, and 
thence show 7 the double values of the sine, the cosecant, the 
cosine, the secant, and the single value of the cotangent. 

7. Show that the f ormulse of the corollary to theor. 4, taken 
two and two, are symmetric : 

those for sine, in terms of cosine, tangent, secant, • • • 

with those for cosine, in terms of sine, cotangent, • • • ; 

those for tangent, in terms of sine, cosine, secant, • • • 

with those for cotangent, in terms of cosine, sine, • • • ; 

those for secant, in terms of sine, cosine, tangent, • • • ^ 
with those for cosecant, in terms of cosine, sine, 

8. Show that the formulas proved in examples 3, 4, page 7, 
hold true with the broader definitions of the trigonometric 
ratios, given on page 34. 

9. Show that the methods of proof shown in examples 5, 6, 
page 7, apply to the formulae in the corollary to theor. 4. 



5, §6.] RATIOS OF RELATED ANGLES. 41 

§ 6. RATIOS OF RELATED ANGLES. 

THE RATIOS OF OPPOSITE ANGLES. 

Theor. 5. If ale any plane angle, then: 
sin ( — <*)= — sin a, cos ( — a)= + cos a, 
tan ( — a) = — tan a, cot ( — a) = — cot a, 
sec [ — a)~ + sec a, esc ( — a) = —esc a. 

For, let xop, xop' be any opposite angles a, —a, having the 
same vertex o, the same initial line ox, and the termi- 
nal lines op, op' symmetric as to ox ; 

draw oy normal to ox and oy' opposite to oy. 

On op, op' take equal segments r, r' and let their major and 
minor projections, i.e. their projections on ox, oy, be 

•&> y> % j y ? 





thenvthe angles xop, p'ox are eqnal, and so are the seg- 
ments r, r', [constr. 
.'. the projections of r, r' on ox are equal, [theor. 1, cor. 3. 
• the angles poy, y'op' are equal, 

. the projection of r on oy equals the projection of r' 
on oy', and is the opposite of the projection of r on 
oy ; [theor. 1, cor. 1. 

r — r', x — x 1 , y— — y\ 
:. sin ( — a), =y'/r r = — y/r, = —sin a, 
cos ( — a), = x'/r' = x/r, = cos a ; 
and so for the rest. 



So 



i.e. 



42 



GENERAL PROPERTIES OF PLANE ANGLES. [II, THS. 



THE RATIOS OF THE COMPLEMENT OF AN ANGLE. 

Theor. 6. If a be any plane angle, then: 

sinco-a—cos a, cosco-a — sin a, tanco-a — cot a, 
esc co-a — sec a, sec co-a — csc a, cot co-a — tana. 

For let xop be any angle a ; draw oy normal to ox, and op' 
making the angle p'oy equal to xop ; 




their. • xop' + p'oy = r, xop + poy = r, p'oy = xop, [constr. 
.*. xop'^co-^^poy. 

On op, op' take r, r equal segments, and let their major and 
minor projections, i.e. their projections on ox, oy, 
be x, y, x', y'j 

then*.' xop, p'oy are equal angles, and so are xop', poy, 

.". the major projection of r equals the minor projection 
of r', 
and the minor projection of r equals the major projection 

of r* ; 
i.e. r=r', x = y', y = x', 

.'. sin co- a, =y'/r'z=x/r, ee cos a ; and so for the rest. 



6, 7, §6.] 



RATIOS OF RELATED ANGLES. 



43 



THE RATIOS OF R + a. 

Theor. 7. If ale any plane angle, and r a rigid angle, 
then: sin(R + a) = cos a, cos(R + a)=z —sina, 

tan(a + a)= —cot a, cot(u + a)= —tail a, 
sec (r + a) = — esc a, csc(u + a) = sec a. 

Let xop be any angle a, draw oy normal to ox, op' normal to 
op, and ox' opposite to ox ; then xop'^r + o-. 

On op, op' take equal segments r, r', and let their major and 
minor projections be x 9 y 9 x' 9 y' ; 

U? AY 



\ p/ 


, 


a 




Vyf 






y 


X 










X' 




y' 

Y N 




i 

i 

1 

i 

! 

1 
1 
1 


X 




' 






X X' 




then*/ xop, yop' are equal angles, 

.\ the projection of r on ox equals that of r' on oy; 
and v poy, p'ox' are equal angles, 

/. the projection of r on oy equals that of r' on ox', and 
is the opposite of the projection of r' on ox ; 
i.e. r=r' 9 x=y r 9 y--x'\ 

.*. sin (n-\-a) 9 =y'/r' = x/r 9 eecos a ; and so for the rest. 



44 GENERAL PROPERTIES OF PLANE ANGLES. [II, TIIS. 

THE RATIOS OF THE SUPPLEMENT OF AN ANGLE. 

Theor. 8. If a he any plane angle, tlien : 

sin sup a — sin a, cos sup a— —cos a, 
tan sup a— —tan a, cot sup a— —cot a, N 

sec sup a — — sec a, esc sup a = esc a. 
Let xop be any plane angle a, draw oy normal to ox, and ox' 

opposite to ox ; 
draw op', making the angle p'ox' equal to a ; 
-p' 




>s? 




t 


J 




PV 




y 


vy^ 




I s 

1 


XT 


V 




1 
1 

<, 1 




^< 


X\ 


i. ,_ 


^ 



X' 


X 


i 



X 


X* 


X 




i 


/**- 


^ 


! 






^/r 


V 


y' 


rX. 


















P 


p 














theny xop' + p'ox' = 2r, and xop = p'ox', 
.•. xop, xop' are supplementary angles. 

On op, op', take equal segments r, r\ and let their major and 
minor projections, i.e. their projections on ox, oy, be 
x, y, x\ y' ; 
then-.* xop, p'ox' are equal angles, 

.*. the projection of r on ox equals that of r' on ox', and 
is the opposite of the projection of r r on ox ; 
and ".' poy, yop' are equal angles, 

.-. the projections of r, r' on oy are equal ; 
i.e. r — r\ x——x\ y = y' ; 

.\ sin sup a, ^y'/r' — y/r, =sin a, 
cos sup a, = x'/r' = — x/r 9 = — cos a ; 
and so for the rest. q.e.d. 



8, 9, §6.] RATIOS OF RELATED ANGLES. 

THE RATIOS OF 2ll + a. 



45 



Theor. 9. If a be any plane angle, and R a right angle, 
then : sin(2n + a) = —sin a, cos(2R + a) = —cos a, 
tan (2r + a) = tan a, cot (2n -r-a) = cot a, 

sec (2r + a) = — sec a, esc (2k + a) = — esc a. 
Let xop be any plane angle a ; draw oy normal to ox, op 
opposite to op ; 




their/ pop' = 2r, 

.*. xop' = 2r + <*. 
On op, op' take equal segments r, r f , and let their major and 
minor projections, i.e. their projections on ox, oy, be 

*> y> x '> y' ; 

then*.* r, r' are opposite segments of op, 

.*. their major projections are opposite, and so are their 
minor projections ; 
i.e. r=r', x=-x', ij=-y'; 

.-.sin (2n + a),==y'/r'= — y/r,=z— sin a, 
cos (2r -f a), =x' /r' — —x/r, ~ — cos a ; 
and so for the rest. Q.E.D. 



46 GENERAL PROPERTIES OF PLANE ANGLES. [II, TH. 

QUESTIONS. 

1. Given the ratios of £r: by aid of theors. 5-9, find the 
ratios of J-r, fR, |r- • •, and of — %n, — §r, — j-B, — Jr • • •. 

2. Given the ratios of £r: find those of f-R, |r, |r • • •, 
and of -Jr, -|r, -|r, -4r - . .. 

3. Given the ratios of R: find those of 0, 2r, 3r, 4r • • •, 
and of — r, — 2r, — 3r, — 4r 

4. Find the ratios of r + a, as the complement of — a. 

5. Find the ratios of 2r + ^, as the supplement of —a. 

6. Find the ratios of a — r, as the opposite of co-a. 

7. Find the ratios of oR + a, and of 3r — a. 

8. Find the ratios of 2r — a, as the complement of a — r. 

9. Find the ratios of 4r + ^ as supplement of — (2n + a). 

10. Given cos a — \ : find sin -1 ^, sin -1 — \. 

11. Given esc a — 2 : find sec -1 2, sec -1 — 2. 

12. What angles have the same sine as a ? the same cosine? 
the same tangent ? the same secant ? the same cosecant ? 

In ratios of positive angles less than e, express the values of : 

13. sin 135°, 335°, ~535°, ~735°, % 7 R, " 5 T 9 ^ ¥ R > "ff R - 

14. cos 235°, 435°, -635°, "835°, ^R, ~V R , -^r, -f£fi. 

15. tan 335°, 535°, "735°, "935°, ^R, ~^R, ^-R, ~ffR. 

16. cot 435°, 635°, ~835°, -1035°, - 8 ^R, ~^r, ^R, "iffc 

17. sec 535°, 735°, ~935°, "1135°, ^r, -y-R, ^R, -f|R. 

18. esc 635°, 835°, "1035°, "1235°, ^r, S&n, ^R, ~ffR. 
In ratios of positive angles not greater than |-r, express the 

values of : 

19. sin 50°, 150°, "250°, "350°, t ^r, ~4r. 

20. cos 60°, 160°, "260°, "360°, T ^R> "^R- 

21. tan 70°, 170°, ~270°, ~370°, j\r, -yR. 

22. cot 80°, 180°, "280°, "380°, t ^r, ~6r. 

23. sec 90°, 190°, "290°, "390°, -i^R, Sfn. 

24. cscl00°, 200°, "300°, "400°, i|r, -^r. 



10, § 7.] PROJECTION OF A BROKEN LINE. 



47 



§7. PROJECTION OF A BROKEN LINE. 

The projection of a broken line upon a straight line is the 
sum of the projections upon it of the segments that consti- 
tute the broken line, and it is identical with the like projec- 
tion of the single segment that reaches from the initial to the 
terminal point of the broken line. 



> 


re 




/L 






Q\ 


sr 


E' 




F 
B" 


B 


/ > 




^ 


Cf 
\ 




r" 


/ 


s^< 







A 1 / 





** 


D' /H 


>^ H \ 

! \ 


X 


TV' 




/ ■" 


>" 




V c ' ] 


?' V 


i 
E L 




F" 




//-' 






f®~y ■ l 




A" 


™ 


A 




N 






P 



Theor. 10. Tlie major projection of a segment of a line is 
equal to the product of the segment by the cosine of the angle 
the line makes with the line of projection ; and the minor pro- 
jection is equal to the product of the segment by the sine of this 
angle. [df. sine, cosine. 

Cor. The major projection of a broken line is the stem of 
the products of the segments each by the cosine of the angle its 
line makes with the line of projection, and the minor projec- 
tion is the sum of their products by the sines of these angles. 
E.g. in the figure above, let the broken line abcdef be 

formed by the segments ab, bo, cd • • •, of the lines 

al, bm, cn • • •, 
let a, (3, y • • • <t> stand for the angles ox-al, ox-bm, ox-cn 

• • -OX-AF. 

maj-proj abcdef = a'b' + b'c' + c'd'- *• 

= AB cos a + BC COS 13 + CD cos y • • • 
= A r F' = AF COS 0, 

min-proj abcdef = a"b" + b"c" + c"d"« • • 

= ab sin a + BC sin f3 + CD sin y • • ■ 
= a"f"=a'e sin (j). 



then 



and 



48 



GENERAL PROPERTIES OF PLANE ANGLES. [II, TH. 



g 8. RATIOS OF THE SUM, AND OF THE DIFFERENCE, 
OF TWO ANGLES. 

Theor. 11. If a, f3 be any hco plane angles, then : 
sin (a + fi) = sin a cos /? + cos a sin /3, 
sin {a — /3) — sin a cos ft — cos a sin /?, 
cos (a + /?) = cos a cos (3 — sin a sin /?, 
cos (a — (3) — cos a cos (3 + sin a sin j3. 

For, let xop, poq. be any two plane angles a, /?, so placed 
that their vertices coincide,, and the terminal line, op, 
of a, is the initial line of f3 \ 

then xoq =:<* + /?. 

On oq take any segment oc, and draw cr normal to op at B ; 



VR 

/ 
/ 
/ 


(c 




-L 

/ 




i** 




then*.' the major projection of oc, as to ox, equals the like 
projection of the broken line obc, 

i.e. maj-proj oc — maj-proj ob + maj-proj bc, [df. 

.'. maj-proj oc/oc = maj-proj OB/oc-f maj-proj bc/oc 
= maj-proj ob/ob • ob/oc + maj-proj bc/bc • bc/oc. 



11, §8.] THE SUM AND DIFFERENCE OF TWO ANGLES. 49 

But maj-proj oc/oc = cos xoq = cos (a + /3), [df . 

maj-proj ob/ob = cos xop= cos a ; 
and v ob, bc are the major and minor projections of oc, as 
to OP, 
.-. ob/oc = cos poq = cos/?, 
bc/oc = sin poq = sin ft ; 
and v angle ox-br = xop -hPBR = ^ + R, 

. \ maj-proj bc/bc = cos ox-br = cos (a + r) = — sin a, [th. 7. 
.*. cos (a + /3) = cos a cos fi — sin a sin /?. Q. E. D. 

And'.* a, /3 may be any plane angles positive, or negative, 
and a — /3=a + ( — /3) whatever the sign or magnitude of fi, 
.'. cos (a — )3) — cos a cos ( — /?) — sin a sin ( — fi) [df . 

= cos a cos ft -Ksin a sin /?. q.e.d. [theor.o. 
So, Y the minor projection of oc equals the like projection 
of the broken line obc, 
.*. min-proj oc/oc = min-proj ob/oc -f min-proj bc/oc 
= min-proj ob/ob • ob/oc + min-proj bc/bc • bc/oc, 

.•. sin (or + /?) = sin a cos /? + sin (r -f «)■ sin /3 

~ sin <* cos /3 + cos a sin /3, q. e. d. 

and sin (a — /3) = sinacos ( — fi) + cos # sin ( — /?), 

= sin or cos /? — cos # sin /?. q.e. d. 

Cor. 1. tan (a + /3) = (tan a + tan y8)/(l — tew <* tew /?), 
tew- (a — /3) — (tan a — tan fi)/(l + tern a tan /?). 

For tan (a + /?) = sin (a -f /?)/cos (<* H- /?) [theor.4. 

= (sin or cos /? + cos a: sin /?)/(cos a? cos /? — sin « sin /?). 

Divide both terms of this fraction by cos a cos ft ; 

then tan (a + /3) = (tan or + tan fi)/(l — tan a tan /?) ; 

and so for tan (a — /3). q.e.d. 

Cor. 2. sin(a-\-/3)+sin (a — /3) = 2sinacos j3, 
sin (a + /?) — s£m (a — /3) = 2 cos a sin fi, 
cos (a + ft) + cos (a — ft) —2 cos a cos /3, 
cos (a + fi) — cos (a — fi) — — 2 sin a sin /?. 



50 GENERAL PROPERTIES OF PLANE ANGLES. [II, TH. 

CONVERSION FORMULA. 

Theor. 12. If a, ft be any tivo plane angles, then : 
sin a + sin ft — 2 sin% (a + ft) cos^(a — ft), 
sin a — sin ft — 2 cos \ (a + ft) sin% (a — ft), 
cos a + cos ft — 2 cos \ (a + ft) cos J (a — ft), 
cos a -cos ft— —2 sin \ (a + ft) sin % (a — ft). 
For let y, d be two plane angles such that 

r=H a +fi) and *?=H a -A)-> 

then, y + d=a, and y — d = ft, 

and V sin (y + d)+sin (y — 6) = 2 sin ^cos S, [theor. 11, cor. 2. 

.". sin or + sin /? = 2 sin J (a + ft) cos ^ (a — ft) ; 
and so for the other formulas, q.e.d. 

questions. 

1. Given sin a—.Z, sin /? = .6 : find sin (a + ft), sin (a — ft), 
cos (a + ft), cos (a — ft), each correct to three decimal places. 

2. From the sine and cosine of 30° and 45°, find the ratios 
of 15° and 75°, then those of 105°, 165°, 195°, 255°, 285°, 345°. 

3. Eemembering the ratios of 0, r, 2r • • •, verify theors. 
5-9, by aid of theor. 11. What is the defect in this proof ? 

4. If a,* ft be any plane angles, then 

sin (a + ft) • sin (a — ft) — sh\ 2 a — sin 2 /? = cos 2 /? — cos 2 **, 
cos (a + ft) • cos (a — ft) — cos 2 a — sin 2 /? = cos 2 /? — sin 2 a\ 

5. Divide the values of sin (a + ft), sin (a — ft), cos (a + ft), 
cos (a — ft), each in turn by cos a cos ft, sin a sin ft, sin a cos ft, 
cos a sin ft, and express the results in terms of tan a, tan ft. 

6. Show that, with a, ft each smaller than two right angles, 
there may be thirty-two distinct figures to illustrate theor. 11, 
each differing from the rest in some important particular. 

E.g. a, ft, a + ft may all be positive acute angles, or a, ft 
may be positive acute angles, and a + ft an obtuse angle. 

7. If sin a = sin/? and cos a— cos ft prove that 

cos (a — ft) = 1, and so that a, ft are either equal or congruent. 

8. Prove that cos a + cos (120° + a) + cos (120°- a) = 0. 



12, § 8.J THE SUM AND DIFFERENCE OF TWO ANGLES. 



51 



9. Prove that sin 2 10° - cos 2 190° = cos 200°. 

10. If A, b, c, d be any four plane angles, then 

sin (a — b) sin (c — d) + sin (b — c) sin (a — d) 
+ sin(c — a) sin(B-D) = 0. 

11. Let a, ft, y, • • • X be any plane angles, then 

cos (a + ft) sin (a — /?)+cos (fi + y) sin (ft — y) + • • • 
-f cos (A, + a) sin (A — a) = 0. 

12. Solve the equation cos 3 a: + cos 2 or + cos or = 0. 

[3a = 2a + a, a — 2a — a. 

13. Prove the identity [7a = 4:a + 3a, 5a = 4:a + a- • • 
sin a -f sin da + sin ha + sin 7^ = 4 sin 4# cos 2a cos a. 

14. Given tan a: = \, tan ft — \\ find tan (# + /?). 

15. Given tan a — \, tan/?=r^, tan/ = ^: find tan (a + ft + y). 

16. Show that sin 28° + sin 14° = 2 sin 21° cos 7°, 

sin 28° - sin 14° = 2 cos 21° sin 7°, 
cos 28° + cos 14° = 2 cos 21° cos 7°, 
cos 28° - cos 14° = - 2 sin 21° sin 7°, 
sin 80° -sin 20° = cos 50°, 
sin 75° -sin 45° = sin 15°. 

17. In terms of tangents and cotangents find the values of: 
(sin a + sin /J)/(cos a + cos ft), 

(sin a — sin /?)/(cos a + cos ft), 
(sin a 4- sin /J)/(cos a — cos ft), 
(sin a — sin /J)/(cos # — cos ft), 
(sin <* + sin /?)/(sin <x — sin /3), 
(cos a + cos /?)/(cos or — cos /J). 

18. Given ar= 60°, /? = 45°: find tan 52° 30'; tan 7° 30'. [th.12. 
So, given ^ = 45°, ft = 30°: find tan 37° 30', tan 7° 30'. 

19. Given sin 15° = .25882, sin 45°=: V^: find cos 60°, cos 30°. 

20. Given cos 75° = .25882 : find sin 30°. 

21. Given cosl7° = .9563, sin 23°==. 3907 : find tan 7°, tan 
40°, sin 20°, cos 3° 30'. 



52 GENERAL PROPERTIES OF PLANE ANGLES. [II, THS. 

§9. RATIOS OF DOUBLE ANGLES AND OF HALF ANGLES. 
Theor. 13. If a he any plane angle, then : 
sin 2a = 2 sin a cos a — 2 tan a/(l + ta?i 2 a), 
cos 2a — cos 2 a — sin 2 a 
= 2cos 2 a-l 
= 1 — 2 sin 2 a 

= (1 - tan 2 a)/(l + tan* a), 
tan 2a~ 2 tan a /(I — tan 2 a). 
For sin 2a = sin (a + a) 

= s'ma cos a + cos a sin a? [theor. 11. 

= 2 sin # cos # ; q.e.d. 

= (2 sin a'/cos a) - cos 2 a: 
= 2 tan a/sec 2 a 

— 2 tan a /{I + tan 2 a). [theor. 4. 
So, cos 2a — cos {a + a) 

= cos a cos # — sin a sin a [theor.ll. 

— o,os 2 a — sin 2 a ; Q.E.D. 
and cos 2a = cos 2 a — (1 — cos 2 ^ ) 

= 2 cos 2 ^ — 1 ; Q. e. d. [theor. 4. 

and cos2tf=(l — sin 2 **)— sm 2 a 

— 1 — 2 sin 2 a; Q.E.D. 
and cos 2a = (cos 2 a/cos 2 a — sm 2 a/cos 2 a) • cos 2 ^ 

= (1 — tan 2 ^)/sec 2 «r 

= (1 — tan^)/(l + tanV). ■ q.e.d. [theor. 4. 

So, tan 2a — tan (a + a) 

= (tan a + ta,na)/(l—tari a tana:), [th.ll, cor.l. 
= 2 tan a /{I — tan 2 a') . Q. e.d. 

Cor. sin a — 2 sin \a cos \a, 

cos a — 2 cos 2 £ar — 1 = 1 — 2 sin* \a, [theor. 4. 

\^cosa — 2cos 2 \a y 
1 — cos a — 2 sin 2 ia. 



13, 14, §9.] RATIOS OF DOUBLE ANGLES AND HALF ANGLES. 53 

Theor. 14. If ale any plane angle, then : 
sin^a— Vi(l— cos a), 
cos \a = */^(l + cosa), 
tan %a = sin a/( 1 + cos a) 

— {1 — cos a) /sin a 

= V [(1 — cos a) /(I + cos a)]. 

For v 2 sin 2 \a — 1 — cos <*, [theor. 13, cor. 

.*. sin^<2= \J\(Y — cos a), q.e.d. 

So, v 2cos 2 -£ar=:l + cosa', [theor. 13, cor, 

.\ cos|of=\/|(l-fcostf'). Q.E.D. 

So, v tan|tf' = sin^<:v/cos^-a', [theor. 4. 

.\ tan \a — 2 sin \a cos \a/2 cos 2 %a 

— sin a /(I + cos a) ; q. e. d. [theor. 13, cor. 
and tan \a — 2 sin 2 \a/2 sin \a cos \a 

= (1 — cos a) /sin a. Q. E. D. 

So, tan \a — sj\{\ — cos a)/ \J\ ( 1 + cos a) 

= V[(l— COS tf)/(l + COStf)]. Q.E.D. 

QUESTIONS. 

1. From the known value of cos 30°, find the ratios of 15°; 
from cos 15° find the ratios of 7° 30'; from cos 7° 30' find the 
ratios of 3° 45', and so on, each correct to four decimal places. 
If a + b + c = 2r, then : 

2. sin 2a + sin 2b + sin 2c = 4 sin A sin b sin c. 

3. sin A + sin b + sin c = 4 cos -J- a cos ^b cos ^c. 

4. cos a + cos B + cos c = 4 sin £a sin ^b sin Jc + 1. 
Prove the identities : 

5. csc2^ + cot2a' = cot a; cos <* = cos 4 ^tf — sin 4 ^. 

6. tsiTia + cot a — 2 esc 2a ; tan a — cot a= — 2 cot 2a. 

7. tan (^R — \a) -f-cot (£r — ^d) — 2 sec a. 

8. (cos a + sin a)/ (cos a — sin a) ~ tan 2a + sec 2a. 

9. tan 2 (^R + \a) — (sec a + tan a) /(sec a — tan a) . 



54. GENERAL PROPERTIES OF PLANE ANGLES. [II, TH. 

§10. KATIOS OP THE SUM OF THREE OR MORE ANGLES, 
AND OF MULTIPLE ANGLES. 

Theor. 15. If a, ft, y be any three plane angles, then : 
sin (a + ft -f y) — sin a cos ft cos y + sin ft cos y cos a 
+ sin y cos a cos ft — sin a sin ft sin y 

— cos a cos ft cos y (tan a + tan ft + tan y — tana tan ft tany). 

cos (a + ft 4- y) — cos a cos ft cos y — gos a sin ft sin y 
— cos ft sin y sin a — cosy sin a sin ft 

— cos a cos ft cosy (1 — tan ft tany — tany tana — tana tan ft). 

( /? , \ _ t an a + t an ft + t an y ~ t an a t an ft t an y 

\ / /)— i — tan ft tany — tany tana — tana tan ft' 
Prove by expanding sin (a + ft + y), cos (<* + /? + ;/). [th. 11. 
Cor. 1. If a, ft, y, • • • be any plane angles, then : 
sin (a + ft + y • • • )/cos a cos ft cos y • • • 

= 2 tan a — 2 tan a tan ft tan y 3- • • • , 
cos(a + ft-\-y- • • )/cos a cos ft cos y->- 

= 1 — 2 tan a tan ft + 2 tan a tan ft tan y tan 8 • • • , 
wherein 2t&na stands for the sum of the tangents of all 
the angles, 2 tan a tan ft for the sum of their products taken 
two and two, and so on. Prove by induction. 

Cor. 2. sin 3a = 3 sin a cos 2 a — sin z a = 3 sin a — 4t sin*a, 
cos 3 a = cos* a — 3 cos a sin 2 a = — 3 cos a + 4 cos 3 a, 
sin 4ia = 4: sin a cos z a — 4 sin z a cos a 
= 4 sin a cos a — S sin 3 a cos a, 
cos 4ta = cos* a — 6 sin 2 a cos- a + sin* a 
= 1 — 8 cos 2 a + 8 cos* a, 
• sinna=nsin a cos n ~ } a 

— \n (n — 1) (n — 2) sin 6 a cos n ~ 3 a + • • • , 
cosna = cos n a — ^n(n — l)sin 2 a cos n ~ 2 a+ • • • , 

wherein the coefficients in the value of sin na are those of the 
second, fourth ••• terms of the expansion of (a + b) n ; and 
those in the value of cos na are the first, third • • • terms of 
the same expansion. Prove by induction. 



15, § 10.] THREE OR MORE ANGLES, MULTIPLE ANGLES. 55 

QUESTIONS. 

1. If a, /?, y be any three plane angles, then : 

— sin {a 4- (3 4- y) + sin ( — a + (3 4- y) + sin (a - (3 4- y) 

4- sin (a + /3 — y) — ^ sin a sin /? sin y. 
cos (or 4- /? 4- y) 4- cos ( — a + /? 4- y) 4- cos (<* — /?+ }/) 
+ cos (or 4- /? — /) = 4 cos or cos /? cos y. 
If A4-b4-c = 2r, then : 

2 . tan ^A tan £b 4- tan ^B t an £c 4- tan ^C tan \ A = 1 . 

3. cot A + cot b 4- cot c = cot a cot b cot c 4- CSC A CSC B CSC c. 

4. tan A 4- tan b 4- tan c = tan A tan b tan c. 

5. tan 3 a = (sin a 4- sin 3a + sin 5 ar)/(cos a 4- cos 3^4- cos 5 a) . 

6. If a, f3 be any two plane angles, and n any integer, then : 

[sin a 4- sin (a + fi) 4- sin (a + 2/3) + sin (a + 3/5) H 

4- sin (a -f n — 1/3)] - 2 sin ±/3 
= cos (a — \j3) —cos (a + ?i — ^/3), 
[cos a + cos (# + /?) 4- cos (or4-2/J)4-cos (a + 3fi)+ • • • 
4- cos (a + n — 1/5)] -2 sin^/? 
= sin (a + n — \fi) — sin (a — ^j3). 

7. From the results of ex. 6, prove that : \n any pos. integer. 

sin a 4- sin («4-4r/w) 4- • • • -fsin [# + 4r(^ — l)/w] = 0, 
cos #4-cos(tf + 4R/ft)4- • • • +cos [#4-4r(^ — l)/n] = 0. 

8. In the results of ex. 7, take n = 3, and prove that : 

sin a 4- sin 60° -a - sin 60° + ^ = 0, 



cos a — cos 60° — a — cos 60° 4- or = 0. 
9. In the results of ex. 7, take n — h, and prove that 



sin a 4- sin 72° + a + sin 36°-^- sin 36° + a- sin 72° -or = 0, 



cos <* 4- cos 72° 4- r*- cos 36° -a: -cos 36° 4- <* 4- cos 72° -a:=0. 

10. Show that when n — 3 the formula found in ex. 7 veri- 
fies the sines and cosines of all angles in the first quarter, if 
to a be given values from 0° to 30°. 

11. In the results of ex. 7, take n — 9, 15, 25, 27, 45, in turn, 
and thence find other formulae of verification. 



56 GENERAL PROPERTIES OF PLANE ANGLES. [II, 

§11. INVERSE FUNCTIONS. 

If a be a number, and a an angle such that a = sin a, this 
relation is expressed by the equation a — sin -1 #, which is read 
a is the anti-sine of a. So, the equation /3 = cos -1 # means 
that J3 is an angle whose cosine is h, and y^tan -1 *?, that y 
is an angle whose tangent is c. 

It is to be noted that, while the equations a = sin a, 1> — cos /?, 
c = tan y, give a, 5, 6? single values for single values of a, /3, y, 
the equations a: = sin -1 a, /? = cos _1 5, y = tnn~ 1 c give a, /3, y 
many values for single values of a, h, c : for a, 2r — a, and all 
the congruents of these angles have the same sine ; /?, — /?, and 
all their congruents have the same cosine ; and y, 2r + /, and 
all their congruents have the same tangent. 

E.g. sin-|=:30 o , 150°, 390°, 510° ... -210°, -330°.... 

So, cos~Vi = 45°, -45°, 315°, -315° .... 

So, tan-y3 = 60°, 240°, 420°, 600°. - • -120°, -300°,. • •• 

Many of the theorems of trigonometry may be expressed in 
terms of inverse functions ; and sometimes with advantage. 

E.g. if x y y, z stand for the sines of the angles a, /?, y, 
then sin (a ± /?) = sin a cos /? ± sin /? cos <*, may be written 

sin~ 1 a?zi:sin"" 1 ^=sin~ 1 [x\/(l — y 2 ) +y*J{l— x 2 )~\. 
So, sin (a + /3 + y) = sin a cos ft cos y + sin ft cos y cos a 

-f sin y cos a cos ft — sin a sin /? sin y, may be written 
sin~ 1 a; + sin~ 1 y + sin~l0 = sin~ 1 [x\f{l— y 2 — z 2 + y 2 z 2 ) 
+y*/ (l-z 2 -x 2 + z 2 x 2 ) +z*J (1-x 2 -y 2 + x 2 y 2 ) -xyz]. 
So, sin 2a — 2 sin <* cos <#, may be written 

2 sin- 1 a? = sin- 1 [2x\/ (1-x 2 )]. 
So, sin^^= Vi (1 — cos #) may be written 
isin- 1 ^=zsin-Vi[l-V(l-^)]. 
These relations are always true : 
sin-^^csc -1 !/^, cos- 1 ^ = sec _1 l/^, tan- 1 ^=cot _1 l/^, 
sin-^ + cos-^^Rj sec- J ^ + csc -1 ^=:R, tan-^ + cot'^^R. 



§11.] INVERSE FUNCTIONS. 57 

QUESTIONS. 

Translate these formulae into inverse forms : 

1. cos (a ±/3) = cos a cos/? + sin a sin/?. 

2. tan(or±/?) = (tano'±:tany5)/(l=ptana' tan/?). 

3. cos 2a = cos 2 ** — sin 2 <* = 2 co$ 2 a — 1 = 1 — 2 sin 2 a\ 

4. tan 2a: = 2 tan a /(I — tan 2 a). 

5. cos \a— Vi(l + cos a). 

6. tan \a — sin a/( 1 -+- cos a) 

— (1 — cos or) /sin a 

= V[(l — cos <*)/(! + cos <*)]. 

7. cos (a + /3 + y) — cos a cos ft cos y — sin # sin /? cos y 

— sin /? sin ;/ cos a — sin y sin a cos /?. 

8. tan (« + /? + >/) 

tan <* + tan /? 4- tan ;/ — tan or tan fi tan y 



1 — tan a tan /? — tan /? tan y — tan ;/ tan a: 
9. cos 3a = cos 3 a: — 3 cos a sin 2 a. 

10. tan 3a = (3 tan or- tan 8 or)/(l -3 tan 2 **). 
Show that 

11. sin~ 1 -g + siii" 1 -|=B; cos~ 1 T ^+cos~ l 4f =b; 

tan _1 f + tan~ 1 f = r. 

12. sin (3 sin -1 #) = 3 x — 4 x z . [x any proper fraction, 
cos (3 cos" 1 a;)= —3 x- j r 4zX z . 

tan (3 tan -1 x) = (3 x — x 3 ) :(1 — 3 x 2 ). [x any number. 

13. tan-^ + tan-^^R. [Euler. 

14. tan -1 -J + tan -1 J- + tan -1 ^=r | r. [Dase. 

15. 2tan- 1 i + tan~ 1 i=:-|R. [Hutton. 

16. 4 tan -1 1 -tan -1 2^-9 = | r. [Machin. 

17. 4tan- l i-tan- l T 1 o+ tan ~ l -g l 9=iK. [Rutherford. 

18. 5 tan- 1 1 + 2 tan- 1 ^9 = ! r. [Euler. 
Solve the equations : _ 

19. sin _1 3ic + sin~ 1 4a:=:R. 

20. tan- 1 2a: + tan- 1 3.r = ^R. 



58 



GEKERAL PROPERTIES OF PLANE ANGLES. 



[n, 



§12. GRAPHIC REPRESENTATION OF TRIGONOMETRIC 

RATIOS. 

Let xp be any arc with centre o and radius ox, and let py be 

the arc complementary to xp ; 
through p, x draw ap, xt normal to ox, and through y, p draw 

by, pt' normal to op, with t on op and t' on oy ; 




then ap, oa, xt, ot are the sine, cosine, tangent, and secant of 

the arc xp ; 
and by, ob, pt', ot' are the sine, cosine, tangent, and secant of 
the complementary arc py, and the cosine, sine, cotan- 
gent, and cosecant of the arc xp. 
These lines are called line-functions of arcs as distinguished 
from the ratio -functions of angles ; and if they be divided by 
the radius, the ratios so found are the ratio-functions hereto- 
fore defined. With arcs of the same radius the ratios of their 
line-functions are equal to the ratios of the like ratio-func- 
tions of their angles. 



§13.] 



GRAPHIC REPRESENTATION OF RATIOS. 



59 



CURVE OF SINES. 

Let ox be the radius of a circle, and divide the circumference 
into any convenient parts at p x , p 2 - • • ; 




draw AxP 1? a 2 p 2 - • • normal to ox, and sines of the arcs xp 2 , 

xp 2 - • •; 
upon ox lay off XB t , xb 2 - • • equal to the arcs xp 1? xp 2 - • • ; 
at b 1} b 2 - • • erect perpendiculars to ox and take c 2 , c 2 - • • such 

that B^i^AxPi, b 2 c 2 = a 2 p 2 • • • ; 
through c 1? c 2 • • • draw a smooth curve ; it is the curve of sines, 

and the following relations are manifest : 
The sine is for the angle ; 
is nearly as long as the arc for a small angle ; 
increases more and more slowly ; 
is equal to the radius, and its ratio is +1, its maximum, for a 

right angle ; 
decreases, at first slowly, but faster and faster as the angle 

approaches two right angles ; 
is for two right angles ; 
decreases from to the opposite of the radius, and its ratio is 

— 1, its minimum, as the angle grows from two right 

angles to three ; 
increases to as the angle grows from three right angles to 

four ; 
is again at the end of the first revolution; and so on. 

The sine has all values between the radius and its opposite. 

If the revolution be continuous, the values of the sine are 

periodic, every successive revolution indicating a new cycle 

and a new wave in the curve. The sines are equal for pairs 

of angles symmetric about the normal at o. 



60 GENERAL PROPERTIES OF PLANE ANGLES. [II, 

OTHER TRIGONOMETRIC CURVES. 

The tangent is for the angle ; 

increases through the first quarter to -f-oo ; leaps to — oo ; 

increases through the second quarter to ; 

increases through the third quarter to + oo ; leaps to — oo ; 

increases through the fourth quarter to ; and so on. 

The tangent has all values from — oo to -f oo . Tangents are 
equal for pairs of angles that differ by a half revolution. 

The secant is equal to the radius, and its ratio is +1 for the 
angle ; 

increases through the first quarter to 4- oo ; leaps to — oo ; 

increases through the second quarter to the opposite of the 
radius, and its ratio is — 1 ; 

decreases through the third quarter to — oo ; leaps to -f oo ; 

decreases through the fourth quarter to the value at the begin- 
ning ; and so on. 
The secant has no value smaller than the radius. Secants 

are equal for pairs of angles symmetric as to the initial line. 

The cosine, cotangent, cosecant have the same bounds as the 
sine, tangent, secant ; they go through like changes and are 
represented by like carves ; but they begin, for the angle 0, 
with different values, viz., the radius, oo, oo. 

QUESTIONS. 

1. Show directly from the definitions what are the largest 
and what the smallest values that each function may have, and 
state for what angles the several functions take these values. 

So, what are the greatest and what the least values. 

2. Draw the curve of tangents, curve of secants, curve of 
cosines, curve of cotangents, and curve of cosecants. 

3. Trace the changes, when a increases from to 4r, in : 

sinaf + coso', tan a: + cot**, sina' + csca', 
sin a — cos a, tan a — cot a, sin a — esc a. 



§12.] GKAPHIC REPRESENTATION OF RATIOS. 61 

QUESTIONS FOR REVIEW. 

1. Find sec (a±ft), esc (a±ft) in the ratios of a and ft. 

2. Given tan 1° 30'= .0262 : find tan 21°, tan 24°, cot 21°, 
cot 24°. 

Show that : 

3. cos 2a: = 2 (sin a + ^n) (sin tf + fR). 

4. sin (fi + y — a) + sin (y + a — ft) + sin {a + ft — y) 

— sin (a + ft + y) = 4 sin a sin ft sin ;/. 
5.. cos 2 (/? — ;/) + cos 2 (;/ — <*) + cos 2 (# — ft) 

= 1 + 2 cos {ft — y) cos (y — a) cos (a — j3). 

6. tan a + tan /? = sin (# + /?)/cos<* cos/5. 

7. tan ^ (a + /?) = (sin a + sin ytf)/(cos # + cos /?). 
Solve these equations : 

8. 4 sin 6> sin 3 = 1. 

9. sin 30 -sin = 0. 

10. tan0 + tan20 = tan30. 

11. cos — sin 0=Vi. 

12. 3cos0 + sin0 = 2. 

Trace the changes in sign and magnitude as grows from to 
4r, in : 

13. cos20/cos0. 

14. sin — sin-J0. 

15. tan + cot 0. 

16. sin + sin 20 + sin 40. 
Prove the equations : 

17. tan- 1 T 2 T + 2tan- 1 i = tan- 1 £. 

18. cot- 1 2 + csc"VlO = iR. 

19. sin- 1 :^ + tan- 1 (l-2:) = 2tan-V(^-^). 

20. sin- 1 [2xy(l+^ 2 )]+tan- 1 [2V(l-^ 2 )] = R. 

21. If tan^0 = tan 3 |^ and tan <^ = 2 tan a, then + (j)—%a. 

22. If sm(x + a)/sm(x + ft) = V (sin 2 a/sin 2/?), 
then tan 2 x = tan a tan ft. 



62 



PLANE TRIANGLES. 



[HI, 



III. PLANE TRIANGLES. 



§1. THE GENERAL TRIANGLE. 

Let a, b, c be any three directed lines that meet each other, 
b, c at a, c, a at b, a, b at c 3 the figure so formed is a plane 
triangle, abc. 




Of the eight figures shown here, the first may be called the 
ideal triangle : its sides, taken in order, and followed in their 
positive directions each till it crosses the next one, form a 



§1.J THE GENERAL TRIANGLE. G3 

closed figure, and the primary angles be, ca, ab are all posi- 
tive. In going round the other figures from vertex to vertex 
in order, some of the sides must be followed in their negative 
directions and some of the angles are negative. Such triangles 
may be called deformed triangles. 

The eight figures show all the possible ways of describing a 
plane triangle : for each side must be traversed in one of two 
ways, forward or backward ; and the two ways of describing 
the first side may be combined at will with the two ways of 
describing the second side, and these 2-2 ways, with the two 
ways of describing the third side, making 2-2-2 ways of de- 
scribing the three sides. 

If a, /3, y stand for the exterior angles of the triangle, i.e. 
for the angles be, ca, at, then, in the ideal triangle, a, /3, y 
are the supplements of the interior angles, a, b, c, commonly 
called the angles of the triangle, and their sum is four right 
angles. In the deformed triangles the sum of a, /3, y is some 
congruent of four right angles. * 

QUESTIONS. 

1. What is the effect on the values of the sides and angles 
of an ideal triangle, of reversing the direction of one of the 
bounding lines ? of reversing two of the bounding lines ? of 
reversing all three of them ? of keeping the directions fixed 
and moving one bounding line parallel to itself, to a position 
equally distant from, and on the other side of, the opposite 
vertex ? of turning over the plane of the triangle ? 

2. If a man, walking around a triangular field, abc, start 
at a, walk to B, turn to the left so as to face c, walk to c, turn 
to the left so as to face A, walk to A, turn to the left so as to 
face b, through what angle has he turned ? 

3. So, if, starting from a and going about the field, he face 
in the direction ba, and walk backward from a to b having 
the field on his right, then, facing in the direction cb, walk 
backward to c, then, facing in the direction AC, walk back- 
ward to A, through what angle has he turned ? 



G4 



PLANE TRIANGLES. 



[Ill, TH. 



§2. GENERAL PROPERTIES OF PLANE TRIANGLES. 

The discussions below apply directly to the ideal triangle, 
but with due attention to signs they apply to the deformed 
triangles as well. 

The letters a, b, c have a double use : first as the names of 
the indefinite directed bounding lines of the triangle, second 
as the segments bc, ca, ab, of these bounding lines. These 
segments are the sides of the triangle. 
E.g. in the statement a is the angle bc the indefinite lines 

b, c are meant and a is their angle ; 
but in the equation a 2 — b 2 + c 2 + 2 5c cos a a, b, c are the 
measured and directed sides. 
The context always shows clearly which use is intended. 

LAW OF COSINES. 

Theor. 1. If a, b, c be the sides of a plane triangle, and 
q, /3, y the angles be, ca, ab, then ; 

a 2 = b 2 + c 2 + 2bc cos a, 

b 2 = c 2 + a 2 + 2 ca cos /3, 

c 2 = a 2 + b 2 + 2abcosy. 
For, project the closed broken line a + b + c on a ; 




thenv /. db—y, Z ac— — /?, 

.*. a + b cos y + c cos ( — /?) = 0, 
i.e. a -f 5 cos y + c cos /3 = 0. 
So, b + c cos a + acosy = 0, 
and c + a cos /3 + b cos a — 0. 



[II, theor. 10. 

[II, theor. 5. 
[project a + b + c on b. 
[project a + b + c on c. 



1, §2.] GENERAL PROPERTIES OF PLANE TRIANGLES. 65 

Multiply the first of these equations by — a, the second by b, 

the third by c\ and add ; 
then — a 2 + b 2 + c 2 + 2bc cosa = 0, i.e. a 2 = b 2 -\- c 2 -\- 2bc cos a ; 
For the second formula multiply by a,—b, c and add, and 
for the third multiply by a, b,—c and add. 

Cor. 1. cos^a- V[(s-S) (s-c)/2bc]. [2s = a + b + c. 

For v 2cos 2 ^a / = l + cos a 

= l + {a 2 -b 2 -c 2 )/2bc 
= (a 2 -b 2 + 2bc-c 2 )/2bc 
= [a 2 -(b-c) 2 ]/2bc 
= (a-b + c)(a + b- c)/2bc 
= ±(s-b)(s-c)/2bc, 
.\ cos ^a=*J[(s — b(s — c)/bc]. q.e.d. 

Cor. 2. sin^a — \J[s (s — a) /be]. 
For v 2 sin 2 ^a = 1 — cos a 

= l-(a 2 -b 2 -c 2 )/2bc 
= (b 2 + 2bc + c 2 -a*)/2bc 
= [(b±c) 2 -a 2 ]/2bc 
= (b + c + a)(b + c-a)/2bc 
= 4cS (s — a)/2bc, 
.'. sin^=: \/[s (s — a) /be], q.e.d. 

Cor. 3. cot\a~ \/[(s-b) (s-c)/s(s-a)]. 
Cor. 4. If a, b, c, a, (3, y be the parts of an ideal triangle, 
and if a, b, c be the interior angles of the triangle, then : 
cosA=(b 2 + c 2 -a 2 )/2bc, 
sin -Ja = V [(s — b) (s — c)/bc] , 
cos |-a = V [s (s — a) /be], 
tan^A.= \/[(s — b) (s — c)/s (s — a)]. 
For '.'a, a are supplementary angles, 

.'. \a, ^-a are complementary angles, 
and cos a —— cos a, cos^ar = sin ^a, 

sin \a — cos \k, cot %a = tan ^a. 
5 



66 



PLANE TRIANGLES. 



[Ill, TH. 



LAW OF SINES'. 

Theor. 2. If a, b, c be the sides of a plane triangle, and 
a, /?, y be the angles be, ca, ab, then : 
a /sin a — b/sin /3 — c/sin y. 

For, draw any normal to a and project the closed broken line 
a + b + c on this normal ; 




thenv Aab—y, lac= — (3, 

and the projection of a on its normal is naught, 

.-. + b sin y + c sin ( - /?) = 0, 

.-. b -sin y = c-sm ft, 

.-. b/sm/3 = c/sin/. 
So, c/sin y = a/sin a, 
and fl/sin # = 5/sin ft. 

Cor. 1. (a + b)/c = cos%(a — /3)/c6s%y. 

(a — b)/c = — sin %(a — f3)/sin \y. 
For '.■ a/c = sin a/sin y> b/c = sin /?/sin y, 
.'.(« + J)/c = (sin or -f sin /?)/sin y 

= 2 sin £(<* -f- /S) cos J(a: - /?)/2 sin £;/ cos^ 
= cos £(a - /?)/cos Jy. [J-(« +/?) = supiy, 

So, (a — b)/c— (sin or — sin /J) /sin y 

= 2 cos £(<* + /?) sin i(^ ~ /^)/2 sin^ cos^-y 
= — sin ±(a — /?)/sin ^y. 

Cor. 2. (a - J)/(a + b) = - tan %(a - fi)/tan\y. 



[project a + J + cona normal to b. 
[project a + J + cona normal to c. 



[above. 



2»§&] GENERAL PROPERTIES OF PLANE TRIANGLES. 67 

Cor. 3. If a, b, c, a, (3, y be the parts of an ideal triangle, 
and if a, b, c be the interior angles of the triangle, then : 
a/ sin a = b/sinB = c/sinc, 
(a + b)/c - cos i(A - B)/sin ic, 
(a — b)/c — sin $ (a — b)/cos \c, 
{a - b)/(a + b)= tan i (a - B)/cot £c. 
For v a, a are supplementary angles, and so are fi, B and y 9 c, 

.-.*(«-/?)= -*(a-b), 
and iy, ^c are complementary, 

/. sin a = sin A, sin/? = sinB, sin y = sine, 

cos \y — sin \q, sin %y = cos -Jc, cot -J- y = tan -Jc, 
tanKo- — /?)= — tan J(a-b). 

questions. 

1. What is the value, in terms of a, b, c, of : 
cos /3, cos i/3, sin \fi, cot ^-y5 ? 

cosb, sin-jB, cos^B, tan-^B? 
cos y, cos \y, sin -j^ c °t iy ? 
cose, sin^c, cos^-c, tan -Jc ? 

2. What signs are to be given to the radicals in theor. 1, 
cors. 1, 2, 3, in case of an ideal triangle ? 

3. Show that the values of cos \a, sin \a, • • • are impos- 
sible if one side be greater than the sum of the other two. 

In any plane triangle abc : 

4. cos^-a cos^B/sin-Jc^s/e. 

5. cos -J A sin ^b/cos -Jc = (s — a)/c. 

6. sin-J-A cos ^-b/cos |c = (s — b)/c. 

7. sin £a sin ^-B/sin ±c=(s — c)/c. 

8. «cosb + Z>cos a = c ; acosB-Jcos A= (a 2 — b 2 )/c. 

9. a cos b cos c 4- b cos c cos A -f c cos A cos b 

= a sin B sin c = b sin c sin A — c sin A sin B. 

10. a cos a + b cos b -f c cos c = 2a sin B sin c = • • • • 

11. a sin (b - c) + 5 sin (c - a) -f 6* sin ( a — b) = 0. 



08 PLANE TRIANGLES. [Ill, PR. 

§3. SOLUTION OF PLANE TRIANGLES. 
PROB. 1. TO SOLVE AN OBLIQUE PLANE TRIANGLE. 

Apply such of the formulce oftheors. 1, 2, and their corollaries, 

as serve to express the values of the unknown parts in 

terms of the knoivn parts. 
Check : Form an equation involving the three computed parts ; 

but use no part in the same way in the solution and the 

check. 
Cases of the general triangle appear in discussing the rela- 
tions of coplanar forces in mechanics, in particular when one 
of the forces is the resultant of the other two ; and in the solu- 
tion of such triangles, the general formulae given above may 
be used. For ordinary purposes the ideal triangle alone is suf- 
ficient, and in its solution it is convenient and in accord with 
usage to ignore the exterior angles a, f3, y, and to use the 
interior angles A, b, c. The rules may then take the form 
shown below. There are four cases. 

(a) Given a, b, c, the three sides : 
then cos A = (b 2 + c 2 — a 2 )/2 be, 
cosB=(c 2 + a 2 -b 2 )/2ca, 
cos C-(a 2 + b 2 -c 2 )/2ab ; check: a + b + c = 2r. 

These formulae are used if a, b, c be expressed in numbers 
so small that the squares, sums, and quotients are easily com- 
puted ; and the angles are then found from their natural co- 
sines. If a, b, c be expressed in large numbers use the formulae 
shown below, which are specially adapted to logarithmic work. 
tan^A= \/[(s — a) (s — b) {s — c)/s]/(s — a), 
tan^B= \/[(s-a) (s-b) (s-c)/s]/(s-b), 
tanjc=: \/'[(s-a) (s-b) (s-c)/s]/(s-c). 

For tan^A= -\/[(s-b) (s-c)/s (s-a)'\ 

= ^[(s-b)(s- c) [s - a)/s [s - a) 2 ] 
= ^[(s-a){s-b) (s-c)/s]/(s-a), 

and so for tan ^B, tan Jc. 



1, §3.] SOLUTION OF PLANE TRIANGLES. GO 

The special advantage of these formulae lies in this, that the 
radical part is the same for each of the three half angles. 

E.g. Let a, b, c be 3, 5, 7 ; then, using the upper formulas,, the 
work may take this form : 

cosa=(25 + 49-9)/70 = 65/70= .9286, and a = 21° 47' 
cosb = (49 + 9-25)/42 = 33/42= .7857, and B= 38°13' 
cosc = (9 + 25-49)/30 = -15/30= -.5000, and c = 120° 
check: A-f-B + c=180°. 180° 

So, let a, b, c be 357, 573, 735 ; then, using the lower formulas, 
the work may take this form : 

357 8 = 832,5 log, 2.9204- 

573 s-« = 475.5 2.6772 + 

735 5-5 = 259.5 2.4141 + 



2) 1666 s-c= 97.5 1.9890 + 

832.5 check: 1665. 2 )4.1599 

2.0800 
2.0800 2.0800 2.0800 

-2.6772 -2.4141 -1.9890 



log-tan |A = 9. 4028 log-tan Jb = 9. 6659 log-tan £c = 0. 0910 
ia = 14°11 / iB = 24°51^ £c= 50° 58' 

A = 28°22' b = 49°43' c = 101°56' 

check: a + b + c = 180° nearly. 

QUESTIONS. 

1. Show by the formulae that a triangle is possible only when 
each side is less than the sum of the other two sides. 

What sign must be given to the radical in an ideal triangle ? 

2. Solve the triangle, given a, 127 m.; 5,64.9 m.; c,152. 16 m. 

[55° 19.4', 24° 51.1', 99° 49.2'. 

3. Solve the triangle, given a, 659.7 ; b, 318.2; c, 527.6. 

4. Solve the triangle, given a, 625 ; b, 615 ; c, 11. 
Before solving show which of the angles a, b, c are large, 

which small, and which smallest. 
Can an exact solution be made ? 



70 PLAXE TRIANGLES, [III, PR. 

(b) Given a, b, c, ttvo angles and a side: 
then c = 180° — (a + b), a — sin A-c/sin c, 5 = sin B-c/sinc. 
check : sin^-c = V[(s — a) (s — b)/ab]. 

E.g. let a, b, c be 50°, 75°, 120 yards ; then the work may take 
this form : 
c = 180°-125 o = 55° 
log 120 =2.0792 2.1658 2.1658 

log-sin 55° = 9. 9134 log-sin 50° = 9. 8843 log-sin 75° = 9. 9849 
2.1658 log a = 2.0501 log 5 = 2.1507 

a = 112.2 yards. b— 141.5 yards. 

check : c =120 

= 112.3 log, 2.0501- 
5 = 141.5 2.1507- 



2)373.7 



5 = 186.85 

s-a= 74.65 1.8730 + 

s-b= 45.35 1.6566 + 

s-c= 66.85 2 )9.3238 ^c = 27 o 30" 

373. 7 9. 6644 log-sin £c =9, 6644 

"questions. 

1. In examples under this case, is there always a solution ? 
Is there ever more than one solution ? What limitations are 
there on the values of the tw r o given angles A, B ? 

2. Solve the triangle, given a, 34°; B, 95°; c, 1389 ft. 

[51°, 9.995, 17.805. 

3. Solve the triangle, given b, 58° 30'; c, 120° 13'; a, 5387 yds. 
Can an exact solution be made with the angles B, c so large, 

and A so small ? 

4. Write out the formulae for the solution and the check 
when b, c, a are given. 

So, when c, A, b are given. 
So, when a, b, a are given. 

5. Why may not more than three parts be given ? 

E.g. Why may not the data be a, 50°; b, 75°; a, 20 ; b, 30 ? 



1, §3.] SOLUTION OF PLANE TRIANGLES. 71 

(6*) Given a, b, c, two sides and their angle : 
then £(a + b) = 90° - Jc, tan | (a - b) = cot jfc • (a - 1) /{a + 5), 

J(A + B)+i(A-B) = A, J(A + B)-J(A-B) = B, 

c = sin c-«/sin a. 

c/j^i : (b + c)/a — cos £( b — c)/sin £a. 

Kg. let a, J, c be 635, 361, 61° 17' ; then the work may take 
this form : 



check , 



FORMULAE. 


NUMBERS. 


LOGARITHMS. 


cot^c 


30° 38' 


0.2275 + 


-a — b 


274 


2.4378 + 


:a+b 


996 


2.9983- 


= tan£(A — b) 


24° 541' 


9.6670 


90° -£c 






= i(A + B) 


59° 21f 




A (sought) 


84° 16 f 




B (sought) 


34° 27' 




a 


635 


2.8028 + 


•sine 


61° 17' 


9.9430 + 


:sin A 


84° 16' 


9.9978- 


= c (sought) 


559.75 


2.7480 


b + c 


920.75 


2.9642 + 


:a 


635 


2.8028- 
0.1614 


= cos^(c-b) 


13° 25' 


9.9880 + 


: sin^-A 


4?° 8' 

QUESTIONS. 


9.8266- 
0.1614 



1. In examples under this case, is there always a solution ? 
Is there ever more than one solution ? Are any limitations to 
be put upon the lengths of the sides or the magnitude of their 
angle? Between what limits do £(a + b), J(a — b) lie? 

2. Solve the triangle, given a, 25.3 ; b, 136 ; c, 98° 15'. 

[10° 10', 71° 35', 141.86. 



72 PLAKE TRIANGLES. [Ill, PR. 

(d) Given b, c, b, two sides and an angle opposite one of them: 
then sin C = c> sin B/b, a = 180° — (b + c), a — sin A • b/sin B. 
check : (a + b)/c = cos ^( A — B)/sin ^c. 

The angle c, found from the equation sin c = b • sin B/b, may 
in general have two supplementary values [sin sup a — sin a, 
and two triangles are then possible. 

But there are some limitations : 

1. If b, b, c be so related that c- sin b > b, then sin c > 1, 
which is impossible, and there is no triangle. 

2. If c>&u\B — b, then sinc = l, c is a right angle, and 
there is one, a right triangle. 

3. If either value of the angle cr makes B > c when b > c } 
or b < c when b < c, that value must be rejected. 

In particular : if b be acute, no triangle is possible if b <p, 
the perpendicular from A to the side a ; one right triangle if 
b—p\ two triangles if p<b<,c; one, an isosceles triangle, if 
b — c\ one triangle if b>c. 

So, if b be right or obtuse, a triangle is possible only when 
b > c, and then but one. 

QUESTIONS. 

1. Draw figures to show the several cases outlined above, 
and show how the geometric constructions interpret the facts 
as shown by the formulae, for the several cases. 

Solve these triangles, given : 

2. b, 18 ; c, 20 ; b, 55° 24'. 

[66° 9', 58° 27', 18.64, or 113° 51', 10° 45', 4.08. 

3. a, 10 ; b, 20 ; a, 30°. 4. b, 16 ; c, 20 ; b, 86° 40'. 
c, 47° 9'. 6. a, 24; 5,20; A, 37° 36'. 
A, 120°. [46° 12', 13° 48', 6.61. 
A, 135°. 9. a, 16 ; b, 20 ; A, 150°. 

10. Let o, p be two points 10 feet apart ; about o describe a 
circle with radius 4 feet ; through p draw a line making an angle 
of 20° with the line PO : at what distance from p does this line 
cut the circle ? 



5. 


c, 20; 


a, 20; 


7. 


a, 24; 


J, 20; 


8. 


a, 20; 


5,20; 



1, §3.] SOLUTION OF PLANE TRIANGLES. 73 

QUESTIONS FOR REVIEW. 

Solve these triangles, given : 

1. a, 40 ; I), 50 ; c, 60. 2. a, 4 ; I, 5 ; c, 6. 

3. a, 411 ; J, 522 ; c, 633. 4. a, 60° ; b, 60° ; c, 10. 

5. a, 24; b, 45° ; c, 24°. 6. A, 31° 26'; 6, 17.1; c, 47° 18'. 

7. a, 14 ; b, 14; c, 60°. 8. a, 38.9 ; b, 9° 18'; c, 119.11. 

9. a, 117° 23'; b,6yc, 11.14. 10. a, 36 ; 5, 40 ; a, 51° 16'. 

11. If the three sides a, h, c of a triangle be given, find the 
length of the perpendiculars from the vertices upon the oppo- 
site sides ; of the lines connecting the vertices with the mid- 
points of the opposite sides ; of the segments of the bisectors 
of the angles, cut off by the opposite sides. 

12. In ex. 10 of page 72, let the distance op be a, the radius of 
the circle I, and the angle poq, c : how many solutions are pos- 
sible when a>b ? when a = b ? when a<b ? 

Show how the angle c is limited in each of these cases. 

13. Discuss ex. 12 if c be negative. So, a or b be negative. 

14. The sides of a triangle are 3, 4, V38 : show, without solv- 
ing, that the largest angle is greater than 120°. 

15. If#, b, c be in arithmetic progression, 3 tan^A-tan^c = l. 

16. If c = 2b, then c 2 = b (a + b). 

17. Show by trigonometry that if an angle of a triangle be 
bisected, the segments of the opposite side are proportional to 
the other two sides. 

18. If a cos a = b cos b, the triangle is either right-angled or 
isosceles. 

19. If p be any point in an equilateral triangle abc, then 
cos (bpc — 60°) = (pb 2 -f pc 2 — pa 2 )/2pb • pc. 

20. Show how to solve a triangle from the three altitudes. 



74 PLANE TRIANGLES. [Ill, 

§ 4. SINES AND TANGENTS OF SMALL ANGLES. 

If an angle be very small, its sine and tangent are also very 
small ; but their logarithms are negative and very large, and 
they change rapidly and at rapidly varying rates. Such loga- 
rithms, therefore, are not convenient for use where interpola- 
tion is necessary, and in their stead the logarithms given below 
may be used ; they are based on the following considerations : 

An angle whose bounding arc is just as long as a radius is 
a radian; it is equal to 57° 17' 44.8", i.e. to 206264.8", and 
the number of seconds in an angle is 206264.8 times the 
number of radians. The index for radians is r . 

For a small angle the number of radians in the bounding 
arc is a very small fraction, and it is a very little larger than 
the sine of the angle and a very little smaller than its tangent : 
it follows that, if a small angle be expressed in radians, the 
ratio sinA r /A is a very little smaller, and the ratio tanA r /A 
is a very little larger, than unity. These ratios approach unity 
closer and closer as the angle grows smaller. 

If the angle be expressed in seconds, the ratio sin a "/a is 
a very little smaller than the reciprocal of 206264.8, and the 
ratio tan a "/a is a very little larger than this reciprocal. These 
ratios change very slowly, and hence interpolation is always 
possible ; the table below gives their logarithms as far as 5°. 

Angle. log(sinA /; /A). Angle. log (tan a' '/a). Angle. log (tan a' 7a). 

0° -1° 18' 

1° 19-1° 59' 

2° -2° 29' 

2° 30'-2° 54' 

2° 55-3° 16' 

3° ir-3° 36' 

The cosine and cotangent of an angle near 90° are the sine 
and tangent of the complementary small angle. The logarithm 
of the cotangent of a small angle is found by subtracting the 
modified logarithm of the tangent of the angle from 10 ; that 
of the tangent of an angle near 90°, by subtracting the modi- 
fied logarithm of the tangent of the complementary small 
angle from 10. 



0° -1° 4' 


4.6856 


1° 5' -2° 23' 


4.6855 


2° 24-3° 11' 


4.6854 


3° 12'-3° 50' 


4.6853 


3° 51'-4° 23' 


4.6852 


40 24'_4o 52 


4.6851 



4.6856 


3° 37-3° 54' 


4.6862 


4.6857 


3°55'-4°ll' 


4.6863 


4.6858 


4° 12 '-4° 27' 


4.6864 


4.6859 


4° 28'-4° 41' 


4.6865 


4.6860 


40 42'_4° 55' 


4.6866 


4.6861 


4° 55-5° 00' 


4.6867 



§4. J SIXES AXD TANGENTS OF SMALL ANGLES. 75 

TO TAKE OUT THE SINE OR TANGENT OF A SMALL ANGLE. 

Take out the logarithm that corresponds to the number of 
degrees and minutes ; and add the logarithm of the whole num- 
ber of seconds in the angle. 
Let A be the number of seconds in an angle ; 
then'.' sin a" = (sin a "/a) • A, 

.'. log-sin a"= log (sin a"/a) + log A ; 
and v tan a" = (tan a"/a) • A, 

.*. log-tan a" = log (tanA"/A) -flog A. 

E.g. log-sin 10' 30" = log (sin 630"/630) +log G30 
= 4.6856* 2. 7993 = 7.4849. 

So, log-tan 3° 13' 40"=log (tan 11620"/11620) +log 11620 

= 4.6860 + 4.0652 = 8.7512. 

The angle is found by a reverse process. 
E.g. to take out log-sin -1 8.4143 : 

From the table of sines and tangents, page xi, it appears 
that the angle sought lies just below 1° 30', and by the formula 

log a = log-sin a" — log (sin a "/a) ; 
and v 8.4143-4.6855 = 3.7288, 

.-. the angle is 5355": i.e. 1° 29' 15". 
So, to take out log-sin" 1 8.8062 : 

The angle sought lies near 3° 40', 
and v 8.8062-4.6853 = 4.1210, 

.-. the angle is 13212"; i.e. 3° 40' 12". 

QUESTIONS. 

1. Find log-sin 22', 43', 1° 11', 1° 27', 2° 24' 36". 

2. Find log-tan 22', 43', 1° 11', 1° 27', 2° 24' 36". 

3. Find log-sin- 1 7.3146, 8.2719, 8.4185, 8.8927. 

4. Find log-tan" 1 7.3146, 8.2719, 8.4185, 8.8927. 
Solve these triangles, given : 

5. a, 327 ; J, 328 ; c, 654. 6. a, 3279 ; 5, 3280 ; c, 1°. 



76 



PLANE TRIANGLES. 



[Ill, TH. 



§5. DIRECTED AREAS. 

If an elastic cord be stretched from a point o to a point a, 
and if while one end of this cord is fixed at o, the other end 
trace a line ab, straight, broken, or curved, the cord, now a 
radius vector of varying length, sweeps over the figure oab, 
and may be said to generate the area oab. It is convenient 
to call the area of the figure oab positive if the radius vector 
oa be positive and swing about o counter-clockwise, and neg- 
ative if it swing clockwise ; and this convention conforms to 
the conventions as to directed lines and angles already in use. 




o A o a o a 

If after generating the area oab the cord swing back from 
OB to oa, and its end retrace the same line from b to A, then 
'the area oab may be thought of as taken up and cancelled, 
and the sum of the areas oab, oba is naught. 
So, if c be any point on the line ab, then : 
area oab + area obc = area oac, 
and area oab + area obc + area oca = 0. 

Theor. 3. If abc be an ideal triangle whose sides are a, b, c, 
arid exterior angles a, ft, y, and if K be the area of this triangle, 




then k = \ab • sin y = \a b • sin c, 

= %a 2 • sin /3 sin y/sin a = ±a 2 • sin B sin c/sin A, 
= *Js (s — a) (s — b)(s — c). 



3, §5.] 



DIRECTED AREAS. 



77 



For draw NA normal to bc, 
their.* k=£bc-ka, 
and KA = CAsin^ 
.-. K=£BC«CAsinj/-, 

i.e. k = \ab sin y — ^ab sine. q.e.d. 

So, V b =asin/?/sina:=:«sinB/sinA, 

.-. K^^a 2 sin/3 sin//sina— ^a 2 sinB sin c/sinA. q.e.d. 

So, V sin ;/ = 2sin^;K cos^y 

= \/s (s — a) (s — b) (s — c)/ab, 
.-. K=\/s(s-a) (s-b) (s-c). Q.E.D. 

Cor. 1. If abc be an ideal triangle, o any point, and K the 
area of abc, then : 

ABC = O AB + OBC + OCA, 

K = 1 [OA • OB Sin AOB + OB • OC si?l BOC + OC • OA si?l COA] . 

(a) o within abc. 




For the three geometric triangles oab, obc, oca are together 
equal to abc, as in the first figure, and their areas are 
all positive. 

(b) o without abc. 

For V when two of the triangles oab, obc, oca are added and 
the third is taken away, the triangle abc remains as 
in the second figure, 

or when from one of them the other two are taken away, 
it remains as in the third figure, 



78 



PLANE TRIANGLES. 



[Ill, THS. 



and while the areas of the two are positive or negative, the 

third is negative or positive, 
,\ the algebraic sum of the areas of these three triangles 

is that of abc, in both cases ; 
/. k = ^[oa-ob sin aob + ob-oc sin BOC + oc • oa sincoA], 




Cor. 2. If abc • • -Lie any polygon, o any point in the plane 
of the polygon, and k the area, then: 

K = £(o a • ob sin aob + ob • oc sin boc H 

-f OL-OA sin LO a). 

In the three theorems that follow, it is assumed that every 
motion of a point is the limit of some motion made up of small 
translations along successive lines, and every motion of a line 
is the limit of some motion made up of small rotations about 
successive points. 

Either motion is that of a point and a line through it, such 
that the point always slides along the line, while the line always 
swings about the point. 

E.g. if a line roll round a circle, without sliding upon it, the 
line always swings about the point of contact, while the 
point of contact always slides along the tangent line. 

The area sivept over by a segment of a straight line is the alge- 
braic sum of the areas of all the infinitesimal quadrilaterals and 
triangles passed over, from instant to instant, by the segment. 



3, 4, §5. J 



DIRECTED AKEAS. 



79 



Theor. 4. If pqr • • • tp be any closed figure traced by the 
end of a radius vector, drawn from o, and varying in lengtli 
if need be, the area of this figure is the area swept over by the 
radius vector, and is positive when the bounding line is traced 
in the positive direction of revolution, and negative ivhen traced 
in the negative direction. 

(a) No reversals of motion of the vector, as in the first figure, 
or only one reversal, as in the second figure : 

For v there are no intermediate reversals, [hyp* 

.*. the figure enclosed by the boundary is swept over once, 
and but once, by the vector, when it swings in the 
direction in which the path is traced ; 

and v all other figures swept over by the vector in one direc- 
tion are also swept over in the other direction, and 
cancelled, 
/. the algebraic sum of the areas of all the figures swept 
over is the area of the figure enclosed by the boundary, 

4 and this area is positive when the path is traced in the posi- 
tive direction of rotation, and negative when it is 
traced in the negative direction. q.e.d. 




o&t. 




(b) Intermediate reversals of motion, as in the third figure : 
For v intermediate reversals occur in consecutive pairs in op- 
posite directions, 

.'.if a point within the enclosure be swept over more than 
once, it is swept over an odd number of times so as 
to give an excess of just one passage in the forward 
direction ; 



80 



PLANE TRIANGLES. 



[Ill, THS. 



and V every point without the enclosure is swept over, if at all, 

the same number of times in each direction, so that 

any outside area that may be generated is cancelled, 

.\ the algebraic sum of the areas of all the figures swept 

over is the area sought. q.e.d. 

Note 1. If the boundary cross itself, the figure is thus di- 
vided into two or more parts : the area of each part may be con- 
sidered separately, and the area of the whole is the algebraic 
sum of the areas of the several parts. 

E.g. the area of the crossed quadrilateral abcd is the algebraic 
sum of the areas of the positive triangle aed and the 
negative angle ebc, and has the sign of the larger. 



d * 




Note 2. In adding two areas any common boundary trav- 
ersed in opposite directions may be erased. 

Cor. If a segment ab of a vector ob swing about o as centre 
■into the position aV, the area of the figure swept over by this 
segment is the area of the figure abb'a', bounded by the initial 
and terminal positions of the segment and the paths of its ends. 




For 



AB = OB - OA, 

the area k of the figure swept over by the segment ab 
is the area of the figure swept over by vector ob less 
the area of the figure swept over by vector oa, 



4, 5, §5.] DIRECTED AREAS. 81 

. \ k = obb' - o a a' = obb' + o a'a = abb'a'a. Q. E. D. [til. 4. 

Theor. 5. If two points k, b move (forward or backward in 
any w T ay) along any paths aa', bb' to a!, b', then the area swept 
over l)\j the straight line ab (varying in length if need be) is the 
area of the figure abb' a'. 

For let the motion of the generator ab be made up of infini- 
tesimal rotations about successive instantaneous centres 
c l9 c i} o 3 • • • ; 




thenv ab sweeps over figures abBxA,, a^b^- • -., [th. 4, cor. 

and v all the intermediate positions A a B 1? a 2 b 2 - • • of ab are 
common boundaries of these figures traced in oppo- 
site directions, 
.*. the sum of all the areas swept over is the area of the 
figure bounded by the path abb'a'a. [th. 4, nt. 2. 

Cor. 1. The area swept over by any straight line ab is the 
sum of the excess of the area of the figure subtended (from any 
origin) by the path of the terminal point b over that subtended 
by the path of the initial point a and the excess of the area of the 
triangle subtended by the initial line ab over that subtended by 
the terminal line a'b'; 




i. e. abb'a'a = (OBB f - oaa') + (oab - oa'b'). 



82 



PLANE TRIANGLES. 



[Ill, THS. 



Cor. 2. If the generator ab return to its initial position, the 
area sivept over is the excess of the area of the figure bounded by 
the path of the terminal point b over that of the figure bounded 
by the path of the initial point a. 





Cor. 3. If the generator ab return to its initial position, 
and the initial point a trace out the same path, to and fro, 
then the area sivept over is the area of the figure bounded by 
the path of the terminal point b. 

Theor. 6. If a wheel be affixed to its axis at the mid-point, 
and if this wheel roll and slide in any way upon a plane tvhile 
its axis remains parallel to the plane, the area sivept over by 
the axis is the product of its length into the distance rolled by 
the wheel. 

For, let ab be the axis and m the mid-point ; 
let the axis turn about an instantaneous centre o, through an 
infinitesimal angle 6, and at the same time let the axis 
slide along its line an infinitesimal distance, to a'b'; 




thenv oa = oa\ ob = ob', om = om', sin#=#, 
.*. area abb'a' = obb' - oaa' = J(ob 2 - oa 2 ) • 6 
=£ (ob-oa) (ob + oa)- = ab-om-0 
= AB-the distance rolled by the wheel at M, 



5, G, §5.] DIRECTED AREAS. 83 

.'. the area swept over by any number of such successive 
rotations is the product of ab by the distance rolled 
by the wheel at M. q.e.d. 

Cor. 1. If the wheel be affixed to its axis at any other yoint, 
c, and the axis turn through an angle, a, between its first and 
last positions, the area stvept over is 

ab • the distance rolled by the wheel at c -f ab • cm • a. 
For v in the infinitesimal rotation above, 
area ab-om- = ab- (oc-f cm)-0 
== ab • the distance rolled by the wheel at c + ab • cm • y 
.-. the sum of all such rotations is AB-the distance rolled 
by the wheel at c + ab-cm-ot. \a— 8+ 0'+ • • •. 

Cor. 2. If the axis return to its first position without making 
a complete revolution, the area swept over is AB-the distance 
rolled by the wheel affixed at any point c. [a = Q. 

QUESTIONS. 

1. If A, b, c be fixed points on a line that turns in a plane 
through an angle a, 

then bc area ab — ab area bc = ^-ab • bc -ca- a. 

2. If the line in ex. 1 return to its first position : 

(a) without making a complete revolution, 
area b = (ab area c + bc area a) : Ac ; 

(b) after making a complete revolution, 

area b + n • ab • bc = (ab area c + bc area a) : AC. 

3. If the chord AC, in ex. 1, slide round an oval, the area 
between the oval and the path of b is n -ab-bc. 

4. Find the area of the curve traced by a point on the con- 
necting rod of a piston and crank in one revolution ; also the 
distance a small wheel attached at the same point would roll 
if a plane surface pressed against it. 



84 PLANE TRIANGLES. [Ill, PR. 

amsler's PLANIMETER. 

Let the axis ab, above noted, be pivoted at a to an arm oa of 
fixed length that turns about a fixed centre o, so that 
A traces a fixed circle while b traces any path whatever ; 
let the wheel be affixed to ab at am T point c, but let it be 
impossible for ab to sweep past oa so that ab, oa can 
take but one position for one position of B, and, if A 
encircle o, ab also encircles o in the same direction : 
1. If a return to its first position without encircling o, 
then*.* a traces out the same path, to and fro, 

.*. the area encircled by b is the area sw r ept over by ab, 

[theor. 5, cor. 3. 
i.e. the area is the product of the number of turns of the 
wheel into the constant area %nr • ab, [theor. 6, cor. 2. 
wherein r is the radius of the wheel. 




2. If A encircle o counter-clockwise, 
then the area encircled by b is the area swept over by AB + the 

area of the circle oa, [theor. 5, cor. 2. 

i.e. the area encircled by b is 27zr- AB-the number of turns 

of the wheel (positive or negative) + ab • cm • a -f n • oa s , 
wherein a is %n. [theor. 6, cor. 1. 

The constants of the planimeter 2nr • ab, 7t(2ab ■ cm -f oa s ) 
can be found once for all. The first is the area due to one turn 
of the wheel ; the second is that due to the swinging of the 
arms oa, ab about o. 



2, §G.] INSCRIBED, ESCRIBED, CIRCUMSCRIBED CIRCLES. 85 

§6. INSCRIBED, ESCRIBED, AND CIRCUMSCRIBED CIRCLES. 

PROB. 2. TO FIND THE RADII OF THE CIRCLES INSCRIBED 
IN, ESCRIBED, AND CIRCUMSCRIBED ABOUT, ANY TRIANGLE. 

For the radius of the inscribed circle, divide the area by half the 

perimeter. 
For the radius of an escribed circle, divide the area by half the 

perimeter less the side beyond ivhicli the circle lies. 
For the radius of the circumscribed circle, divide half of either 

side by the sine of the opposite angle. 
For, let abc be any triangle, and let r = radius of inscribed 

circle, r' , r" , r'" = radii of escribed circles whose centres 

are o', o", o'", and r = radius of circumscribed circle ; 







w 



then*.' k = \r {a + b + c) = rs, [geom. 

,'.r=K/s. Q.E.D. 

So, v K = N £r' ( - a + b 4- c) = r' (s - a), [geom. 

.*. r' = K/(s-a) ; and so for r", r" f . q.e.d. 

Checks : 1/r = 1/r' + 1/r" + \/r m , k 2 = r . r 1 • r" . r"'. 



SG PLANE TRIANGLES. [Ill, PR. 

About abc draw a circle and draw ca', a diameter ; join a'b ; 



thenv A — a', and angle abc is a right angle, [geom. 

and ca' = a/sin a' = a/sin A, 

.'. R = Tj-fl/sin A • • •. Q.E.D. 

Note, a/sin A = #/sinB = c/sinc=:2R. 

QUESTIONS. 

Find the radii of the inscribed, escribed, and circumscribed 
circles, and check the work, given : 

1. a, 12.7; b, 22,8; c, 51.5. 

2. A, 64° 19' ; B, 100° 2' ; c, 51.25. 

3. a, 136 ; b, 95.2; c, 11° 37'. 

4. In a right triangle, 2n + r = s. 

5. If R = 2r, the triangle is equilateral. 

6. In the ambiguous case the two values of R are equal. 

7. The distances from the centre of the inscribed circle to 
the centres of the three escribed circles are equal to 

4r sin ^-a • • • , and to a sec Ja • • • • 

8. The square of the distance between the centres of the 
inscribed and circumscribed circles is R a — 2r?\ 

Prove the equations : 

9. r =(s — a) tan ^A. 

10. r — s tan ^A tan ^B tan Jc. 

11. r = a/(cot £B + cot£c), r' = a/(tan£B + tan£c). 



2, § G.] INSCRIBED, ESCRIBED, CIRCUMSCRIBED CIRCLES. 87 

Prove the equations : 

12. n = aI)c/±K. 

13. R = s/(sin A + shiB + sin c). 

14. r' + r" + r'" - r = 4r j rr'/r'Y" = tan 2 Ja. 

15. K = 4Rr COS^-A COS^B cos|c. 

16. R + r = R (cos a + cosb + cosc). 

17. 4r sin a sin b sin c = a cos A + 1 cos B + c cos c. 

18. In the figure on page 85 co'" is perpendicular to o'o", 

AO' to oV", BO" to 0"'o'. 

The point o, the co-point of these three perpendiculars, is the 
orthocentre of the triangle o'o'V". 

The triangle abc, whose sides join the feet of the perpendic- 
ulars two and two, is the pedal triangle of oVo"'. 

19. The circle circumscribed about abc passes through the 
mid-points of the triangle o'o'V, and through the mid-points 
of the segments oo', oo", oo'". 

This circle is the nine-point circle of the triangle o'o"o'". 

20. The nine-point circle of a triangle circumscribes its 
pedal triangle, passes through the mid-point of each side, and 
bisects the lines joining the vertices to the orthocentre. 

21. If a, h, c be the sides of a triangle, and p be the radius of 
the circle inscribed in a triangle whose sides are b + c, c + a, 
a + b, then p 2 = 2nr. 

22. If a, l, c be the sides of a triangle, and m, n, p be the 
altitudes, then mnp = (a + i -f cfr z /abc. 

23. If u, v, iv be the distances between the excentres of a 
triangle, then uvw sin A sin b sin c — §r'r"r' n . 

24. Find the radii of the circles that touch two sides of a 
triangle and the inscribed circle. 

So, of those that touch the circumscribed circle. 

25. Find the relation which exists between the angles of a 
triangle whose orthocentre lies on the inscribed circle. 



88 DERIVATIVES, SERIES, AND TABLES. [IV, 

IV. DERIVATIVES, SERIES, AND TABLES. 



§1. CIRCULAR MEASURE OF ANGLES. 

In the applications of trigonometry to numerical problems, 
e.g. the solution of triangles, the most convenient unit of an- 
gular measure is the right angle, or the degree, the ninetieth 
part of a right angle ; but in certain other problems, e.g. the 
computation of trigonometric ratios and their logarithms, that 
angle which lies at the centre of a circle, and whose bounding 
arc is just as long as the radius of the circle, is a better unit. 
This unit angle is called a radian, and its magnitude is inde- 
pendent of the length of the radius. [geom. 




Radians may be indicated by the sign r , just as degrees, min- 
utes, and seconds are indicated by the signs °, ; , " ; and since 
the ratio of the half circumference of a circle to its radius is 
7T, [3.141592- • •] and angles at the centre are proportional to 
their arcs, two right angles are equal to n radians. 

The primary equation expressing the relation between de- 
grees and radians is 7r r =:180 o : from this it follows that 

17^=90°, 1^ = 45°, -^=30°, •••, 

I r =180°/;r = 57 o 17' 44.8", 

I°=:7r7l80 = .0174533 r , l' = .0002909 r , 
and the measure of other angles is expressed by the ratio of 
the bounding arc to the radius. 



§1.] CIRCULAR MEASURE OF ANGLES. SO 

QUESTIONS. 

1. Prove that the number of radians in an angle is ex- 
pressed by the ratio of the arc subtending it to the radius of 
the circle, i.e. by the number of radii in the arc. 

2. Express in degree-measure the angles : 

\n, \n, \n, \n y 3.1416'', .7854 r , V\ 1.5 r , "2 r , (7i' + l) r . 

3. Express in radius-measure the angles : 

14°, 15°, 24°, 120°, 137° 15', "4800°, 13', 24". 

4. If the radius be an inch, find the length of the arcs : 
14°, 15°, 120°, 57° 17' 44.8", 1°, \n y \7t, 2 r , 7r + l r . 

So, if the radius be five inches. 

5. How many radii in an arc of : 20°, 180°, 3 r ? 

6. If the radius be 10 inches, find the number of radians 
subtended by an arc of : 13 inches, n inches, 10°, 5' 13", three 
quadrants. 

7. The angle 3.42 r is subtended by an arc of 5.71 inches: 
find the radius ; the arcs opposite ^n r , V, 5° ; the angles in 
radians and in degrees opposite a one-inch arc, a two-radius 
arc, five quadrants. 

8. If the circumference of a circle be 30 inches, find the 
arcs opposite n r , 30°, 3 r . 

9. How many radians and how many degrees are subtended, 
by : 2J radius arcs, n radius arcs, 3^ quadrants ? 

10. How many radians in 17° 13' 15" ? in 10° ? 

11. An angle of three radians at the centre of a sphere sub- 
tends a two-foot arc of a great circle : find the radius. 

12. The apparent diameter of the sun, as seen from the 
earth, is half a degree ; a planet crosses the sun's disk in a 
straight line at a distance from its centre equal to three-fifths 
of the sun's diameter : show that the angle subtended at the 
earth by the part of the planet's path projected on the sun is 
7r r /450. 



90 DERIVATIVES, SERIES, AND TABLES. [IV,TH.1,2, 

§2. DERIVATIVES OF TRIGONOMETRIC RATIOS. 

Theor. 1. If 8 be the circular measure of a positive acute 
angle, then sin8<8<tan 8. 

For, let xor be any positive acute angle ; with o as centre, and 

any radius ox, describe a circle cutting op in p ; 
through P, x draw normals to ox, cutting ox in a and op in t ; 

T 




O AX 

their. * ap < xp < xt, [geom. 

and ap/ox = sin 8, xp/ox = 8, xt/ox = tan 8, 

:. sin 8 < 8 < tan 8. q.e.d. 

Cor. 1. If 8 approach zero, the ratios 6 /sin 8, 8 /tan 8 
approach unity. 

For v sin 8 < 8 < tan 8, [above. 

/. l<0/sin 8 < sec 8, [div. by sin#. 

and cos 6 < /9/tan 8 < 1 ; [mult, by cos#. 

and v cos 8 = 1, and sec 8 = 1, when# = 0, 
• .-. 0/sin 8 = 1, and 0/tan 8 = 1, when 8 = 0. 

For definition of limit, derivative, etc., and for proof of the 
necessary properties of limits and derivatives, see any good 
work on the differential calculus. 

Some of the fundamental properties of derivatives are, for 
convenience of reference, set down here as lemmas without 
proof ; they are given in two forms : 

If u, v be functions of any variable x, then : 
Lem. 1. d x (u + v) = d x u + d x v, 

d(u + Y) = du + Sv. 



LM. 1-G,§ 2. j DERIVATIVES OF TRIGONOMETRIC RATIOS. 91 

LEM. 2. D x (u-V)=V-D x U + U-D x V, 

d(u-v)=v-(?u + u-<?V, 

LEM. 3. D x (u/v) = ( V • D X U - U • D x v)/V 2 , 

6 (u/v) = (v • 6 u - u . Sy)/y*. 
Lem. 4. D x u n = wu n " l -D x u, du n = nxj nl -du. 
Lem. 5. d x log Q u = d x u/u, dlog e Jj = Su/\j. 

Lem. 6. If v be a function of v, <md? v « function ofx, then : 

D X U = D V LJ- D X V. 

wherein d x = x-derivative of, 6 = a very small increment of, 
and the sign =, read approaches, means that the difference of 
the two members is infinitesimal as to either of them. 

DERIVATIVES OF THE RATIOS. 

Theor. 2. If 6 he the circular measure of any plane angle, 
then : 

Vq sin — cos 0, Bq esc = —cot 6 esc 6, 
i>0 cos 6— — sin 0, t>q sec = tan sec 0, 
Dq tan 8 — sec 2 , VQeot = — esc 2 0. 
For, let 0' be an infinitesimal angle, the increment of ; 
then*.' sin (0 + 0') -sin = 2 cos (0 + £0') sin %6', 

.-. [sin (0+ 0') -sin 0]/0' = cos (0 + J0') -sin^/^'. 
But V 0' is the increment of 0, [hjT» 

and sin (0-h 6') — sin is the consequent increment of sin 0, 
.*. lim (incsin 0/inc 0), = DQsin 0,= cos 0. q.e.d. [th. 1. 
So, v cos (0 + 0') -cos 0= -2 sin (0 + ^0') sin ^0', 

.-. [cos (0+0') -cos 0]/0'= -sin (0 + ^0') • sin £0'/£0', 
. *. lim(inc cos 0/inc 0), = Dqcos 0,= —sin 0. q.e.d. [th. 1. 
So, Bq tan 0, = d g (sin 0/cos 0), 

= (cos Dq sin — sin Dq cos 0)/cos 2 
= (cos 2 + sin 2 0)/cos 2 = sec 2 0. 
So, for Dfl esc 0, = d (1/sin 0), for D e sec 0, for Dq cot 0. 



92 



DERIVATIVES,, SERIES, AND TABLES. [IV, THS. 



GEOMETRIC PROOF. 

Let 0-xp be any circle, and Q a point on this circle near p ; 

bisect arc pq at r, and join ox, op, oq, or ; 

draw ap, BQ normal to ox ; 

join p, Q, and through P draw a parallel to ox meeting bq in D ; 

let #=Zxop, #'eeZpoq, (0-\-^0')= Zxor, r = radius of circle ; 




O B A X 

then-.- sin d-A.v/r, sin (0 + ff) = bq/V, 0' = arc PQ/V, 
and Zdqp = Zxor, [geom. 

.-. [sin (d+ 0') -sin 0]/0' = DQ/arc PQ 
= (DQ/chord pq) • (chord PQ/arc pq) 
= cos (9 +%&*)• (chord PQ/arc pq), 
/. lim (inc sin '/inc 6), = Dq sin 0,= cos ; q.e.d. [th. 1. 
and so for i>q cos 0, Dg tan • • • . 

DERIVATIVES OF ANTIFUNCTIONS. 



Theor, 3. d x sin' 

D x COS 

D x /a;^ 

D x C0t~ ] 
D x S6C" 
D x CSC 

For let — sin -1 ^; 

then*.* sin — x, 

.-. d x sin 0, = cos#-D x #, =1, 
.:. d x = 1/cos = 1/V(1 -^ 2 ) ; 

and so for the rest. 



■ 1 x=i/*/(i-x*), 

•V^-i/v(l-^ 

*«=-l/(l+3fy 

l x—l/x\J{x i — \), 
l x— — l/x\/(x* — l). 



Q.E.D. 



2,3,§2.] DERIVATIVES OF TRIGONOMETRIC RATIOS. 93 

Note. When x stands for sin 8 or cos 8, x may have any 
value positive or negative not larger than unity ; when x 
stands for tan 8 or cot 8, x may have any value whatever ; and 
when x stands for sec 8 or esc 8, x may have any value not 
smaller than unity : for if, in the formulae above, x exceeds 
the bounds named, the function is imaginary. 

QUESTIONS. 

1. If 8 be any plane angle and 8 r be the increment of 8, then : 

inc 2 sin 8= - (2 sin ^8J sin (8 + 8'), 

inc 2 cos 8= - (2 sin \8J cos (8 + 8'), 

inc 4 sin 8= (2sm±8')\ sin (0 + 20'), 

inc 4 cos 8= (2sin±d'ycos(8 + 28'), 
wherein inc* sin # = the increment of the increment of sin 8, 
i.e. [sin (8 + 28') -sin (8+8')] -[sin (8 + 8') -sin 8], 
or sin (8 + 28') -2 sin (8+ 8') + sin 8 ; 
and inc 4 sin # = inc inc inc inc sin 8, 
i.e. sm(8 + ±8')-4:sm(8 + 38') + 6sm(8 + 28') 

-4sin(0+0') + sin0. i 

2. If 6a, 6b, 6c, 6 a, 6b be any simultaneous small changes 
in the values of a, b, c, A, b, that are consistent with the known 
relations of the parts of a right triangle 

[a + b = 90°, a* + b 2 = c 2 , a = csmA, b = ccosa], 
then #b=— 6a, 6c=a/c-6a + b/c- 6b = s'm A- 6a + cos a- 6b, < 

6a=sinA'6c + c cos a • 6a, 6b = cos a • 6c — c sin a • 6 a, 
and [eliminate 6c from the last two equations] 

dA=cos a/c • 6a — sin a/c • 6b = (b6a — a6b)/(a* + b*). 

3. If, in a right triangle, only the values of a, b be given, 
and if these have the possible errors ± a' , ± b'; i.e. if a may pos- 
sibly differ from its assumed value by either + a' or ~al , and b 
by either + b' or ~b'\ show from ex.2 that the resulting values 
of c, A will have the possible errors 

± (aa! + bb')/c = ± (a' sin a + V cos a), [a', V positive, 
and ± (aV + ba')/c 2 — ± (b' sin A + a' cos a)/c. 



94 DERIVATIVES, SERIES, AND TABLES. [IV,LM.7-11, 

So, if only b, c be given, with the possible errors ± b' y ± c\ 
find the possible errors of the other sides and angles. 

So, if only b, A be given, or only c, A, with the possible errors 
*b', ± A / , or ± c\ ± a'. 

4. From the known relations of the parts of an oblique tri- 
angle [a + b + c = 180°, a sin b = b sin A, • • • ] prove that 

(a) <?a + <?b + (?c = 0, 

(b) b cos A • 6 a — a cos b • 6b — sin b • 6a + sin a • 6b = 0, 
c cos b • 6b — b cos c • 6c — sin c • 6b + sin B • 6c = 0, 
a cos c • dc — c cos a • 6a — sin a • 6c + sin c • 6a = 0. 

From these equations, by elimination and reduction, derive 

(c) b-6c + ccosA-6B — sin A • Sc + sin c • da= 0, 
c-B6 + b cos A • 6c — sin a • 6b 4- sin b • 6a = 0, 

with four symmetric equations ; and 

(c?) 5 sin c • 6a — (?a + cos c • 6b -f cos b • dc = 0, 
with two symmetric equations. 

5. If in an oblique triangle only a, B, c be given, and if their 
possible errors be ± «/10000, ± 10 ;/ , ± 15 // , find the possible errors 
of A [ex. 4, a]; ofb [ex. 4, c]; of c [ex. 4, c]. 

Find the values of these possible errors when ABC is very 
nearly equilateral, 5000 feet on each side. 

6. Given the values of c, a, b, with the possible errors ± c', 
± a', ± b\ find the possible errors of b, A, c [ex. 4, c y d\ 

7. Given a, a, b, with the possible errors ± a', *#', ± b f } find 
the possible errors of b [ex. 4, b]; of c, c. 

8. Given a, b, b, with possible errors ± a', ± b / , *Z>', find the 
possible errors of c, a, c : first, when, as in all the above cases, 
the computation is assumed to be exact ; second, when c, a, c 
have the further possible errors ± c" , ± a" , ± c" from decimal fig- 
ures omitted in the computation. 

9. Given a, b, c, with possible errors *#', ± b\ ± c f : find the 
possible error of A, with a possible error in computation of ± a". 



Til. 4, §3.] EXPANSION OF TRIGONOMETRIC RATIOS. 95 

§3. EXPANSION OF TRIGONOMETRIC RATIOS. 

In the expansion of trigonometric ratios the following prop- 
erties of series are made use of : they are all proved in works 
on algebra, and are quoted here for convenient reference. 

Lem. 7. If, after a given term, the terms of a series form a 
decreasing geometric progression, the series is convergent. 

Lem. 8. If one series ~be convergent, and if the term.s of 
another series be not larger than the corresponding terms of the 
first series, the second series is convergent. 

Le:m. 9. If, after a given term, the ratio of each term of a 
series to the term before it be smaller than some fixed number 
that is itself smaller than unity, the series is convergent. 

Cor. The series a + A^-f a 2 £ 2 + a 3 £ 3 + • • • is convergent for 
all values of x that make the limit of the ratio of the (n + l) th 
term to the n th term smaller than unity when n becomes very 
great. 

Lem. 10. If in the series a + a^ + a 2 x 2 + a b x 3 + • • • , the limit 
of the ratio of the (n + l) th term to the n th term, for any value of 
x, be smaller than unity, then, in the derivative series a 1 + 2a 2 # 
+ 3a 3 £ 2 + • • > , the limit of the (n + l) th term to the n th term, for 
this value of x, is smaller than unity, and this series is con- 
vergent. 

Lem. 11. If fouo series arranged to rising powers of any 
same variable be equal for all values of the variable that make 
them both convergent, the coefficients of like poivers of the vari- 
able are equal. 

Theor. 4. If 6 be the circular measure of any plane angle, 
then : 

sin 6=0-6*/3 ! + 8 5 /5 ! - &/1 !+•••, 
cos d = l-6 2 /2 ! + # 4 /4 !-# 6 /6 !+••.. 

For assume sin # = a + a 1 # + a 2 # 2 + a 3 # 3 + . . . ? wherein the a's 
are unknown but constant, and has such values as 
make the series convergent, 



96 DERIVATIVES, SERIES, AND TABLES. [IV, TH. 

and find the first two ^-derivatives of both members of the 
equation ; 

then cos 6 = A, 4- 2a,0 + 3 a 3 # 2 H , 

and — sin 8 = 2 a 2 + 2 • 3 A 3 # -i > 

i.e. sin 6— -2a 2 -2-3a 3 #; 

and both of these derivative series are convergent, [lem 9, cor. 

Take as one of the values of ; 

then*.* sin = and A + AiO + a 2 2 -f a 3 3 -f • • • = a 0? 

.\Ao=0. 

So, vcosO = l and a 1 + 2a 2 + 3a 3 2 + • • • =A lf 

.\A*=1. 

So, VAo + A^ + A^ + A^-f • • • = -2a 2 -2.3a 3 6>-3.4aA- 
• • • for all values of 6 that make both series conver- 
gent, 

.*. a = — 2a 2 , a 2 = 3-4a 4 , a 4 = 5-6a 6 --«, 
and a 1 =-2-3a 3 , a 3 =-4-5a 5 , a 5 = — 6-7a 7 - • •; [lem. 11. 

.'. A , A 2 , A 4 , A 6 , A 8 , •••=0, 

and Ax = l, a 8 =-1/3!, a 5 = 1/5!, A f =-l/7---; 

/. sm^=^-^/3! + ^ 5 /5!-^/7!+---, 
and cos0=l-^/2! + 0*/4!-0y6!+-.-. 

Note. These series are convergent for all finite values of 6. 
For the ratios of successive terms, in that for the sine, are 

07(2.8), 07(4-5), 07(6- 7).- -J 
and, in that for the cosine, 

07(1-2), 07(3-4), 07(5-6), •••; 
i.e. series of fractions such that the limit of the (n + l) ih term 
to the n th term is smaller than unity whatever be the value of 0. 

But they converge rapidly only when 6 is quite small. 

Cob. 1. tan6>=0+073 + 2<97(3-5) + 1707(3 2 -5-7) 
+ 626<7(3 4 -5-7)+---, 
cot0 = l/0-0/3-07(3 2 .5)-207(3 8 -5-7) 
-0V(3 3 -5 2 -7) , 



4, §3.] EXPANSION OF TRIGONOMETRIC RATIOS. 97 

sec^=l + <9 2 /2 + 5(9 4 /(2 3 .3) + 61^/(2^3 8 .5) 

+ 277# 8 /(2 7 .3 2 .7)+--., 
csc0 = 1/0 + 0/(2. 3) + 7# 3 /(2 3 .3 2 . 5) 

+ 310 5 /(2 4 .3 3 .5.7) + 127<9 7 /(2 7 .3 8 .5.7) + .... 

For the tangent, divide the series for the sine by that for the 

cosine ; 
for the cotangent, divide the series for the cosine by that for 

the sine ; 
for the secant, divide unity by the series for the cosine ; 
for the cosecant, divide unity by the series for the sine. 

Note. The series for tan and sec are convergent only 
when < \n, for tan and sec are finite and continuous 
functions of for all values of smaller than ^n ; but when 
6 = j7t their values are infinite. So, the series for cot and 
6 esc are convergent only when <7t. 

Cor. 2. log-sin 6= log 0-0 2 /(2.3)-0 4 /(2 2 .3 2 -5) 
-0 6 /(3*.5.7)-..., 
log-cos 6= -[0 2 /2 + 4 /(2 2 .3) + 0y(3 2 .5) 
+ 17# 8 /(2 3 .3 2 .5.7) + ...]. 
For -Be log-sin = cos 0/sin = cot = 1/0 - 0/3 

-0 3 /(3 2 .5) , [lem. 

.\ log-sin = log0-0 2 /(2.3)-0 4 /(2 2 .3 2 - 5) 

-0 6 /(3 4 .5.7) , [lem. 

i.e. . log-sin 6 — the series whose 0-derivative is the above 
series for cot 0, and which, as = 0, approaches to 
log as log-sin must do. 

So, VDfllog-cos0= -sin0/cos0=-tan0= -(0+0 3 /3+ ■ • •), 
.-. log-cos 0= - [0 2 /2 + 4 /(2> • 3) + 6 /(3 2 • 5) 
+ 170 8 /(2 3 .3 2 .5.7) + -..]. 

Note. The series for log-sin is convergent for all values 
of smaller than 7t ; that for log-cos for all values smaller 
than \n. 



98 derivatives, series, and tables. [iv, th. 5, 

Gregory's theorem. 
Theor. 5. If x be any number smaller than unity, then 

tan^x^x-^ + \x* -\x 7 + ^x 9 - T \x ll -{ . 

For, assume tan" 1 ^=A + A 1 a; + A 2 i« 2 + A 3 ^ 3 -l-A 4 ^ 4 + A5^ 5 + • • • , 

and take the ^-derivative of both members ; 
then D a .tan _1 cc=:A 1 + 2A 2 2; + 3A3^ 2 + 4A4a; 3 + 5A 5 ^ 4 4- • • •; 
and '.'V x ttm- 1 x = l/(l+x 2 ) = l-x' 2 + x*-x' + •••., [theor. 3. 

.'. A 1 + 2A a £ + 3A 8 2: 2 + 4A 4 3 8 + • . • =l-X*-\-X i -X 6 +' • ., 

for all values of x that make both series convergent, 
.\ A 2 , a 4 , A 6 ,...=0, [lem.ll. 

and A^l, a 3 = -1/3, a 6 = 1/5, a 7 = -1/7 « • •. 

So, take as a value of x, 

then tan- 1 = A + A 1 + A 2 2 + A 3 3 +. ••, and A = 0; 
.*. t&n.- l x = x — %x z + \x h — \x 7 +\x 9 — ■ • •. q.e.d. 

Note. This series is convergent when #<1; but it con- 
verges very slowly when x is near one, and rapidly only when 
x is small. 

In the computation of the length of an arc, and so of the 
circumference of a circle and of n, either of the equations below 
gives a practical working rule : 

\n- tan- 1 1/V3= [1-1/(3- 3) + 1/(5- 3 2 )-l/(7-3 8 ) 

+ 1/(9 • 3 4 ) - 1/(11 - 3 5 ) + 1/(13 • 3 6 ) ]/ V3. 

±tc = tan- 1 1/2 + tan- 1 1/3 

= 1/2-1/(3- 2 3 ) + l/(5.2 5 )-l/(7.2 7 ) + ... 
+ l/3-l/(3.3 8 )+l/(5.3 5 )-l/(7.3 7 )+--. 
i;r = 4tan- 1 1/5 -tan- 1 1/239, 

= 4 tan- 1 1/5 -tan- 1 1/70 + tan- 1 1/99 

= 4[l/5-l/(3.5 3 ) + l/(5.5 5 )-l/(7.5 7 )+---] 
~[l/70-l/(3.70 3 ) + l/(5.70 5 )-l/(7.70 7 ) + .^] 
+ [1/99-1/(3- 99 3 )+ 1/(5- 99 6 ) -1/(7- 99 7 ) + ...]. 



PR.1,§4.] COMPUTATION OF TRIGONOMETRIC RATIOS. 



99 



With the last of these equations, the work of computation 
may take this form : 



5 


4. 




+ 


- 


25 


.8 


1 


.8 




25 


.032 


3 




.010 666 667 


25 


.00128 


5 


.000 256 




25 


.000 0512 


7 

4 




.000 007 314 


25 


.000 002 0-48 


9 


.000 000 228 






.000 000 082 


11 




.000 000 007 


70 


1. 








70 


.01-4 285 714 


1 




.014 285 714 


70 


.000 204 082 










.000 002 915 


3 


.000 000 972 




99 


1. 








99 


.010101010 


1 


.010101010 


- 


99 


.000102 030 










.000 001030 


3 




.000 000 343 




.810 358 210 


.024 960 045 






7Z-- 


.024 960 045 






.785 398165 
4 




.*. 


= 3.141592 6 


to eight figures. 



§4. COMPUTATION OF TRIGONOMETRIC RATIOS. 
PROB. 1. To COMPUTE A TABLE OF SINES AND COSINES. 

(a) For angles 0° • • • 30° : 
Eeplace 8 by V, 2', 3'- • -in the formulae of theor. 4. 
E.g. v 1' = tt/(180 ■ 60) = 3. 141 592 653 589 793/10 800 
= .000 290 888 208 666, 
.-. sin 1' = . 000 290 888 208 666 -.000 290 888 208 666 3 /3 ! + 
= .000 290 8882; 
cos 1' = 1 - .000 290 888 208 666 2 /2 !+ • • • 
= .999 999 9577; 



100 DERIVATIVES, SERIES, AND TABLES. [IV, PR. 

sin 2' = 2 x .000 290 888 208 666 

-2 8 x .000 290 888 208 666 8 /3 !+ . • . 
= .000 5817764; 

cos 2' = 1 - 2 2 x . 000 290 888 208 666 2 /2 ! + • • • 
= .999 999 8308. 

Note 1. The fraction tt/10 800 once raised to the required 
powers, first, second, third- • •, and divided by the factorials 
1 !, 2 !, 3 ! • • • , thereafter only simple multiples of the quo- 
tients are used. [A small table of these powers and quotients, 
correct to twenty decimal places, is given on page 38 of Jones' 
Six-place Tables, and a larger table in Callet's Tables de Loga- 
rithmes.] At first but two terms of the series are needed ; but 
later, when 6 is larger and the series therefore converges less 
rapidly, and at critical points, e. g. the finding of the value 
of .485795000 + , correct to five figures, more terms must be 
taken. 
Kg. for 30°, = -j7T = .5236O nearly ; 

and sin 30° = .52360-.5236 3 /6 + .5236 5 /120 

= . 52360-. 02392 + .00033-. 00000+ . . - 
= .5, the true value, within less than .00005 ; 
i.e. by the use of three terms of the series, the sine is found 
correct to four decimal places, the same degree of accuracy as 
that assumed for the value of n. 

Note 2. The method shown above may serve whether indi- 
vidual ratios be sought or an entire table ; but if a table, then 
the following method may also be used. 

Assume sin 1' as differing insensibly from arc 1', 
i.e. that sin 1' = .000 290 8882, 
hence, that cos 1', = Vl-shrT, = .999 999 9577 ; 
then in the formulae 

sin (0+ 0') = 2 sin d cos 0'-sin (0- 6'), [II, th. 11, cr. 2. 

cos (6+0') = 2 cos 6 cos 0'-cos(0-0'), 
replace 6 by 1', 2', 3' • • • in turn, and 6' by 1'. 



1, §4.] COMPUTATION OF TRIGONOMETRIC RATIOS. 101 

E.g. sin 2' = 2 sin 1' cos 1' — sinO' 

= 2 x .000 290 8882 x .990 999 9577 - 

= .000 581 I ; 04 x(l-. 000 000 0423) 

= .000 5817764; 
and sin 3' = 2 sin 2' cosl' — sinl' 

= 2 x . 000 581 7764 x (1 - .000 000 0423) 
-.000 290 8882 

= .000 872 6646. 
So, cos2' = 2cosl' cosl' — cosO' 

= 2 x .999 999 9577 x (1 - .000 000 0423) - 1 

= .999 999 8308; 
and cos3' = 2cos2' cosl' — cosl' 

= 2 x . 999 999 8308 x (1 - .000 000 0423) 
-.999 999 9577 

= .999 999 6193. 

(i) For angles 30° ...45°: 
Keplace 0' by 1', 2'. 3'- • • in the formulas 

sin (30° + 0') = cos 0' - sin (30° -ff), [ad. th.,sin30° = +. 
cos (30° + 0') = cos (30° - 0') - sin 0\ 
E.g. sin 30° 1' = cos 1' - sin 29° 59' 

= .999 999 .499 75 = . 500 25, 

and sin 30° 2' = cos 2' - sin 29° 58' 

= .999 999 .499 50 = . 500 50. 

So, cos 30° 1' = cos 29° 59'- sin 1' 

= .866 17-. 00029 = . 865 88, 
and cos30 o 2' = cos29 o 58 , -sin2' = .S65 73. 
(c) For angles 45° • • -90°: apply the formula* 

sin (45° -f 0') = cos (45° - 0'), [II, theor. 6= 

cos (45° + #') = sin (45°*- 0'). 
E.g. sin45 o l , = cos44°59' = .707 31, 
cos45°l' = sin44 c 59' = .706 90. 



102 DERIVATIVES, SERIES, AND TABLES. [IV, PRS. 

VERIFICATION. 

Note 3. The results are tested in many ways : 
(a)'.' sin£0= VK 1-008 0)> cos h^- VK 1 + cos 0), 

/.from cos 45°, = Vi, are found in succession the sines 
and cosines of 22° 30', 11° 15'- • • . 
So, from cos 30°, = ^^3, are found in succession the sines 
and cosines of 15°, 7° 30'. . .. 

(b)'.' sin 26 = 2 sin 6 cos 6, [II, theor. 13. 

cos 30 = 4 cos 3 — 3 cos 6, [II, theor. 15, cor. 

and sin 3G° = cos 54°, [II, theor. 6. 

.*. 2 sin 18° cos 18° = 4 cos 3 18° - 3 cos 18°, 
.-. 2 sin 18° = 4(1- sin 2 18°) - 3, 
.\sinl8° = £(V5-l), cos18° = JV(10 + 2V5); 
thence, in turn, the sines and cosines of 9°, 4° 30', 2° 15'- • •. 

(c) From cos36°, = cos 2 18°-sin 2 18° = i (Vo + 1), 

and sin36 o , = V(l-cos 2 36 o ) = iV(10-2V5), 

are found the sine and cosine of (36° — 30°), i.e. of 6°, thence 
in turn the sine and cosine of 3°, 1° 30', 45', 

(d) From sin (36° + ff) - sin (36° - ff), = 2 cos 36° sin ff 

= <KV5 + l)sin#', 
subtract sin (72° + ff) - sin (72° - ff), = 2 cos 72° sin ff 

= £(V5-l)sin0'; 
then, sin (36° + 0') -sin (38° -0') 

= sin (72° + ff) - sin (72° - ff) + sin ff : 
a formula that serves to test the sines of all angles from 0° to 
90°, if to ff be given the different values from 0° to 18°. 
For other test formulae, see exs. 7-11, page 55. 

PROB. 2. To COMPUTE TABLES OF NATURAL TANGENTS, CO- 
TANGENTS, SECANTS, AND COSECANTS. 

Divide the sines of the angles, in turn, by the cosines ; the co- 
sines by the sines ; 1 by the cosines ; 1 by the sines : 
or, replace 6 by V, 2', 3' • • • in the formula of theor. 4, cor. 1. 



2,3, §4.] COMPUTATION OF TRIGONOMETRIC RATIOS. 103 

PltOB. 3. TO COMPUTE TABLES OF LOGARITHMIC FUNCTIONS. 

From a table of logarithms of numbers take out the logarithms 

of the natural sines and cosines : 
or, replace 6 by 1' 9 %' > 3', • • • in the formulae of th. 4, cor. 2. 

Subtract the logarithmic cosines from the logarithmic sines ; 
the logarithmic sines from the logarithmic cosines ; 
the logarithmic cosines and sines from 0. 

THE METHOD OF DIFFERENCES. 

A more rapid method is this : 

Take out the functions of three, four, or more angles at regular 
intervals, and find their several " orders of differences "; 

by the algebraic " method of differences," find the successive 
terms of the series of logarithms ; 

interpolate for other angles lying betiveen those of the series, 
and verify at intervals by direct computation. 

For safety, four-place tables must be computed to six places ; 
five-place tables to seven places, and so on. 

When the terms of any order of differences are constant, or 
differ very little, the rule that follows may be applied to form 
new terms of the series. 

Add the constant difference to the last difference of the next 
lower order, that sum to the last difference of the next 
loiver order, and so on till a term of the series is reached. 

In the example that follows, the numbers below the rules are 
got by successive addition : 

ANGLE. LOG-SINE. FIKST DIF. SECOND DIF. THIRD DIF. 

18° 9.489 9824 

18° 10' 9.493 8513 ™ -378 

18° 20' 9.497 6824 ttill -371 1 

Q 7Q40 , - *y 

18° 30' 9.5014764 J^|^ -364 * 

~18°40' 9.505 2340 yf b -357 7 

18° 50' 9.508 9559 l co l -350 7 

19° 9.512 6428 3 6869 



104 



SPACE TRIGONOMETRY. 



[V, TH. 



V. SPACE TRIGONOMETRY. 



Space Trigonometry treats of the relations of the parts of 
triedral angles. It is based on the geometry of space, and on 
the principles established in plane trigonometry. 

§1. DIRECTED PLANES. 

A directed plane was defined on page 25. Such a plane may 
be generated by a straight line swinging about another straight 
line that meets it at right angles, in either of two directions. 
E.g. let oz be any straight line, and let ox, perpendicular to 

oz, swing about oz and take in succession the positions 

ox, oy, ox', oy', ox ; 




then ox generates a plane perpendicular to oz. [geom. 

The fixed line about which the other swings is the axis of the 
plane ; and if this axis be so directed that its positive end is 
in front of the plane, it is a normal to the plane. 
E.g. in the figure above, oz is normal to the plane generated 
by ox, but oz' is contra-normal to this plane. 

The plane is then also said to be normal to the line. 
E.g. the plane of the equator is normal to the earth's axis. 



i, §1-3 



DIRECTED PLACES. 



105 



It will be convenient, in this book, to indicate a directed 
plane by naming two directed lines of the plane in such order 
that the least rotation about their co-point, from the line first 
named to the other, generates a positive angle. 
E.g. if I, m be two directed lines that meet, the plane Im is a 

directed plane ; 
and the plane ml is a directed plane that coincides with Im in 

position, but has the contrary direction. 
So the plane xop is a directed plane in which positive rotation 

is from ox to op, by the shortest way. 

p 





The direction of a plane may also be shown by an arrow. 

Theor. 1. Three straight lines meeting at a point, and each 
perpendicular to the other two, may be so directed that each is 
normal to the plane of the other two taken in order. 




E.g. let ox, ot, oz be three directed lines such that ox is per- 
pendicular to oy, oz, and normal to the plane yoz, 
that oy is perpendicular to oz, ox, and normal to zox, 
and that oz is perpendicular to ox, oy, and normal to xoy. 



106 SPACE TRIGONOMETRY. [V. 

QUESTIONS. 

1. If a rod project above a horizontal plane in a direction 
parallel to the earth's axis, in what direction will its shadow 
on the plane swing in the northern hemisphere ? in the south- 
ern hemisphere ? 

So upon a vertical plane ? In what order will the numbers 
be placed on a horizontal sun-dial ? on a vertical sun-dial ? 

2. To an observer standing behind the transparent dial of 
a tower clock, what is the direction of rotation of the clock 
hands ? is it the same for all four faces ? is the actual direc- 
tion of rotation the same in two opposite faces ? 

3. What is a right-hand screw ? 

4. In turning on the nuts that keep the wheels of a carriage 
upon the axles, is the motion clockwise or counter-clockwise ? 
is it the same motion on both sides of the carriage ? 

5. As a carriage is driven forward, how do the wheels turn, 
to one standing on the right side of the roadway ? to one 
standing on the left side ? 

6. If when a carriage is driven forward the rotation of the 
wheels be positive, what is the rotation when the carriage is 
backing ? 

7. If a carriage drive past, on which side of the roadway 
must one stand that the normal to the plane of rotation of the 
wheels may reach towards him ? away from him ? 

8. How must a line of shafting be directed so that it shall be 
normal to the pulleys that are fixed upon and revolve with it ? 

9. If two wheels with parallel axes be so geared that they 
revolve in opposite directions, what relation have the normals 
to their planes of rotation ? 

10. In the figure of theor. 1, how may a point be placed so 
as to be 

in front of all the planes xoy, yoz, zox ? 
in front of xoy, yoz, and back of zox ? 
What other positions may a point have ? 



§2.] 



DIEDRAL ANGLES. 



107 



§2. DIEDRAL ANGLES. 

If two directed planes meet in a directed line, their co-line, 
and one of them, the initial plane, swing about this co-line till 
it coincides with the other, the terminal plane, both in position 
and direction, the diedral angle so generated is the angle of the 
tivo planes. 

This angle- is directed and measured by the plane angle that 
is generated by a normal to the co-line of the two planes, lying- 
in the initial plane and carried by this plane as it swings about 
the co-line till it becomes normal to the co-line in the terminal 
plane. The co-line may be directed at pleasure, but however 
it is directed the plane of the swinging normal must be taken 
normal to this line. 




E.g. let the directed, planes a, b meet in the directed line x'x, 
and let a'a, b'b be normal, in a, b, to x'x at o ; 

then the diedral angle ab is directed and measured by the plane 
angle aob, in the plane normal to x'x at o. 

So, if the directed co-line be xx' ; 

then aa', bb' are normal in a, b, to xx', at o, the diedral angle 
ab is directed and measured by a'ob', in the plane nor- 
mal to xx' at o. 



108 



SPACE TRIGONOMETRY. 



[V, TIL 



It is to be noted that the angle a'ob' as seen from x' is the 
opposite of aob as seen from x, and that the angle bo a' is the 
opposite of boa ; i.e. a reversal of the co-line of the two planes 
reverses their ano-le. 




Theor. 2. The angle of two directed planes is equal to the 
angle of their normals, as seen from the positive end of the di- 
rected co-line of the tivo planes. 

For, in the figure above, draw op, oq normal to the planes a, b ; 
then*.' op is normal to a'a in the plane aob, and oq to b'b, 
.*. the angle poq is equal in magnitude to the angle aob ; 

[geom. 

and v these angles have the same direction in the same plane, 

and the plane angle aob directs and measures the diedral 

angle ab, 

.\ the angle of the two planes is equal to the angle of their 

normals. q.e.d. 

Cor. 1. If the angle ab be a positive right angle, so is the 
angle po a ; op lies in the plane b and coincides with ob, and oq 
lies in the plane a and coincides with oa'. 



2, §2.] DIEDRAL ANGLES. 109 

Note. The student of the geometry and trigonometry of 
space must train himself to see his figures as figures in space,, 
though shown only by diagrams on a flat surface. For the 
most part these diagrams are made up of straight lines and 
curves, and when he looks at the points and lines of his dia- 
grams, he must see the points, lines, and surfaces in space which 
they represent. It will help him to do this if he will close one 
eye and, without moving his head, look steadily at his diagram 
with the other eye : presently it will stand out. 

It will help him, also, if he will hold some object, his book 
for example, or a card, or a wire cage, between the light and 
the wall : he will learn that the shadows are the pictures, pro- 
jections, of his space figures on a plane. Among other things, 
he will see that right angles are rarely projected into right 
angles, that circles are commonly projected into ellipses and 
sometimes into straight lines, and that lines of the same length 
are often unequal ; and he will learn to look back from the 
picture to the figure in space. 

E.g. in the diagram on page 104, the horizontal circle seems 
to be but half as broad as it is long, and the right angles xoy, 
yoz are drawn as angles of 60°, while the right angle zox is 
drawn as an angle of 120° and appears to be the sum of the 
other two. 

So, in the figure on page 108, there are three non co-planar 
•straight lines a'a, b'b, x'x, that meet in a point o and deter- 
mine three planes that meet in the same point. Three circles 
lie in these planes and have o as their common centre ; and 
these circles determine a sphere whose centre is 0. 

To make this figure stand out more clearly arcs that lie on 
the front of the sphere are shown by full lines, while those that 
are behind either of the other planes are shown by broken 
lines ; and so for the diameters. 

The front edge of the horizontal circle is tipped down, while 
the normal op is tipped forward and does not show its full 
length. 



110 



SPACE TRIGONOMETRY. 



[V, TH. 



§3. PROJECTIONS. 

The projection of a point on a line was defined on page 31. 

The projection of a point on a plane is the foot of the perpen- 
dicular from the point to the plane. 

The projection of a directed line on a plane is the co-line of 
the given plane and a plane perpendicular to it through the 
projected line. 

The plane of projection is that plane on which the projec- 
tion is made, the perpendicular plane is the projecting plane, 
and the co-line of the two planes is the line of projection. 

The angle of a line and a plane is the angle of the line of 
projection on the plane, when directed, and the given line. 






I 


B --*~ 


A„ 








I 








1 


I' 




—\""" 


J- 




B' 7 ^ 


-~ A' 


m 




B' 



The projection of a segment of a directed line on a plane, or 
on another directed line, is the segment of the line of projec- 
tion that reaches from the projection of the initial point of the 
given segment to that of the terminal point. It is a positive 
segment if it reach forward, in the direction of the line of 
projection, a negative segment if it reach backward. The 
projection of a broken line upon a directed line is the sum of 
the like projections of the segments that constitute the broken 
line, and it is equal to the projection of the segment that 
reaches from the first initial to the last terminal point. 

The angle of two directed lines that do not meet is that of 
any two lines parallel to the given lines that meet and reach 
forward in the same directions as the lines. 



3, §3.] 



PROJECTIONS. 



Ill 



Theor. 3. If a segment of a directed line be projected on 
another directed line, the projection is equal to the product of 
the segment by the cosine o£ the angle of the two lines. 

(a) The two lines co-planar. [II, theor. 10. 

(b) The two lines not co-planar. 

For, let I, m be two directed lines not co-planar, and let ab be 

a segment of I, and a'b' be its projection on m ; 
through a' draw I' a, line parallel to I and like directed, and 
through A, B draw planes perpendicular to the line m ; 
then'.' these planes are parallel, and a'b", ab are segments of 
parallel lines cut off by parallel planes, 
.*. a'b":=ab; [geom. 

and \ * angle I'm — angle lm, [df . ang. two lines, 

and a'b' = a'b" cos I'm, [II, theor. 10. 

.*. a'b' = ab cosZm. Q.E.D. 

Cor. The projection of a broken line upon a directed line is 
the sum of the products of the segments that constitute the 
broken line by the cosines of their angles with the line of pro- 
jection. 




E.g. in the figure above, let ox, oy, oz be three directed lines, 
each normal to the plane of the other two ; 

let p be any point in space, and project p on the plane oxy at 
b, and B on ox at a ; 

then the projection of the broken line oabp on op is op, 

and op = o a cos xop + ab cos yop + bp cos zop. 



112 SPACE TRIGONOMETRY. [V, TH. 

§4. TRIEDRAL ANGLES AND SPHERICAL TRIANGLES. 

If three planes meet at a point, they form a triedral angle. 
The three face angles and the three diedrals of a triedral are 
its six parts. 

If three directed lines be given that meet at a point, they 
may be taken in such order and their three co-planes may be 
so directed that all the parts of the triedral shall be positive 
and less than two right angles ; and so, if three directed planes 
be given, their co-lines may be so taken and directed that all 
the parts shall be positive and less than two right angles. 
E.g. if boc, coa, aob be three planes whose directed 

co-lines are oa, ob, oc, 
and if these three planes be so directed that the three face 

angles boc, coa, aob, and the three diedrals 

coa-aob, aob-boc, boc-coa are all positive and 

less than two right angles ; 
then the triedral o-abc may be called the ideal triedral of 

the points A, b, c, as to the centre o. 




The three directed planes of a triedral boc, coa, aob 
may be named by the three Eoman letters a, h, c, and so may 
the three plane angles boc, coa, aob j and the three 
diedrals coa-aob, aob-boc, boc-coa by the three Greek 
letters a, /?, y, and so may their three co-lines oa, ob, oc. 



4,§4.] TRIEDRAL ANGLES AND SPHERICAL TRIANGLES. 113 
POLAR TRIEDRALS. 

If through any point normals be drawn to the three faces of 
a triedral, these normals lie,, two and two, in planes perpen- 
dicular to the three edges of the triedral [geom.], and if these 
new planes be so directed that they are normal to the edges of 
the first triedral, a new triedral is formed so related to the other 
that the edges of either of them are normal to the faces of the 
other. Two such triedrals form a pair of polar triedrals. The 
simplest case of such a pair of triedrals is where the six planes 
all pass through the same point. 




Theor. 4. In any pair of polar ■' triedrals, the face angles of 
one of them are equal to the diedrals of the other. 
For the angle of a pair of directed planes is equal to that of 
their normals. [theor. 2. 

8 



114 



SPACE TRIGONOMETRY. 



[V, 



SPHERICAL TRIANGLES. 

If any point be taken as the centre of rotation of a directed 
plane,, and a sphere be described about this point as centre, the 
co-line of the plane and sphere is a circle of rotation of the 
plane, and so it is a directed great circle of the sphere that has 
the same direction as the plane. 

If any diameter of this circle be directed, the tangent at its 
positive end reaching forward in the direction of the circle is 
normal to the diameter, and that at its negative end is contra- 
normal. That diameter of the sphere which is normal to the 
plane is the axis of the great circle, and its ends are the posi- 
tive and negative poles of this circle. 

E.g. the earth's north and south poles are the positive and 
negative poles of the plane of the equator. 





If two directed planes pass through the centre of a sphere, 
they cut it in two directed great circles ; and if their co-diam- 
eter be directed, tangents at its positive end that reach for- 
ward in the direction of the circles are normal to this diameter, 
and their angle, in a plane facing the positive end of the diam- 
eter, measures the diedral angle of the planes. So the tan- 
gents at the negative end of this diameter are contra-normal, 
and their angle is equal to the other in a plane facing the same 
way. The angle of the axes of the two circles is equal to that 
of the two planes. [theor. 2. 

E.g. the angle between the plane of the equator and that of 
the ecliptic, both west-to-east planes, is 23° 27'. 



§•!.] TRIEDRAL ANGLES AND SPHERICAL TRIANGLES. 115 

If about the vertex of a triedral angle as centre, a sphere be 
described, the co-lines of this sphere with the three directed 
planes are three directed great circles, and together they form 
a spherical triangle whose sides subtend the face angles and 
whose angles, when viewed from the positive ends of the edges, 
measure the diedrals of the triedral. The sides meet on the 
co-diameters of the great circles, i.e. on the edges of the trie- 
dral, and these points are the vertices of the triangle. 




If two polar triedrals have a common vertex, and a sphere be 
described about this vertex as centre, the six directed circles 
cut from the six directed planes by this sphere form &pair of 
polar spherical triangles, such that the vertices of the one are 
the positive poles of the sides of the other, and the sides of the 
one, measuring the face angles of the triedral, are equal to the 
angles of the other. 

In this figure, the line oa is normal to the plane b'oc', ob 
to c'oa', oc to a'ob' ; oa' to boc, ob' to coa, oc' to AOB, 



116 SPACE TRIGONOMETRY. [V, 

THE SIXTY-FOUR TRIEDRALS OF THREE CO-POINTAR LINES. 

If a'a, b'b, c'c be three diameters of a sphere that do not 
lie in the same plane, each of these lines may have either of 
two directions. It follows that either a or a' may be taken as 
the positive end of the diameter a'a, and so for B, b' and for 
c, c', and that there may be eight distinct sets of three points 
on the surface of the sphere : 

A, B, C, A', B, C, A, b', C, A, B, c', 

a', b', c, a', b, c', a 5 b', q\ a', b', c', 

i.e. that these three diameters form eight distinct spherical 
triangles, and eight distinct triedrals, in the geometric sense. 




So, each of the three planes of these three diameters, taken 
two and two, may have either of two directions, and the tri- 
angle of one set of points may have eight distinct forms. 

Sixty-four triedrals and sixty-four spherical triangles are 
thus formed with the same three diameters of a sphere, whose 
sides are all positive and less than four right angles, and 
whose angles may be positive or negative. 

These triangles are called the primary triangles, and other 
triangles congruent with these may be formed by adding mul- 
tiples of four right angles to either angle, or one or more great 
circles to either side. 



§ L] TRIEDRAL ANGLES AND SPHERICAL TRIANGLES. 117 











118 



SPACE TRIGONOMETRY. 



[V, LM. 



§5. GENERAL PROPERTIES OF TRIEDRAL ANGLES. 

Lem. 1. If at any point of an edge of a triedral a normal he 
draivn to the opposite face, and if through this normal a plane be 
drawn normal to another edge, the co-lines of this plane with the 
planes adjacent to the edge are perpendicular to the edge. [geom. 

If these lines be so directed that they are normal to the edge, 
each in its own plane, the angle of these tivo normals is equal to 
the diedral of the two planes. [df . ang. of two planes. 

The normal first drawn is normal to that one of the two nor- 
mals ivhicli lies in the opposite face. 




E.g. let o-ABC be a triedral angle, through D any point on the 
edge oa draw ed normal to the opposite face boc, 

and through ed, draw planes normal to the edges ob, oc, cut- 
ting ob in f, and oc in g ; 

then the lines df, fe are perpendicular to ob, and eg, gd to oc : 

and if df, fe be directed normal to ob, and eg, gd to oc ; 

then the plane angle df-fe is equal to the diedral aob-boc, 
and eg-gd to boc-coa. [df. 

So the line ed, drawn normal to the face boc, is normal to the 
line fe in the plane efd and to the line eg in ged. 



1, §5. J GENERAL PROPERTIES OF TRIEDRAL ANGLES. 119 











120 



SPACE TRIGONOMETRY. 



[V, PR. 



To make clear the relations of the parts of the figures on 
page 118., construct a space model as follows : 

Use card-board or stiff paper, and with any centre o and any- 
convenient radius draw a circle ; 

draw the radii oa, on, oc, oa', making the angles aob, boc, 
coa' equal to the given face angles c, a, b ; 

on oa, oa' take od, od' equal, and draw df, gd' normal to ob, 
oc at f, G, and meeting each other in e ; 

cut out the figure, and fold along ob, oc ; 

bring oa, oa' together, and join e, d with a thread : 

ed is normal to the plane boc and. to the lines fe, eg. 



---.o 




The right triangles fed, egd are shown in the figure as 
hinged at fe, eg, and folded down into the plane of the draw- 
ing. The point D is shown at h, h'. These triangles turn up 
when the two faces aob, coa' are turned up, and with them 
they form a solid figure. 

Of the six figures on page 118, the upper middle figure is 
a space figure, the lower middle figure shows the base of this 
figure in its own plane, and the right triangles, at the right and 
left, ftre the right triangles of the sjDace figure, each shown of 
its true size and in its own plane. 

The eight figures on page 119 show the eight spherical tri- 
angles of page 117, with the lines ed, df, fe, eg, gd drawn as 
in the figure on page 118. The reader will note the directions 
of these lines, and the consequent directions of the diedrals 
a, /?, y. The lemma applies to all the figures alike. 



1, §0.] GRAPHIC SOLUTION OF TRIEDRAL ANGLES. 121 

§6. GRAPHIC SOLUTION OF TRIEDRAL ANGLES. 

By a graphic solution is meant a geometrical construction 
of the required figure, such that the parts sought are deter- 
mined and shown without the use of algebraic formulae and 
without computation. Such a solution, often useful of itself 
and quickly made, serves also as an effective check on the 
results of numerical computation. 

Prob. 1. Given three parts of a triedral angle, to 
construct the other three parts : 

(a) Given the three face angles, a, b, c ; 

Through any point of the plane of the paper draw rays oa, 
ob, oc, oa', making the angles aob, boc, coa' equal to 
the given face angles c, a, b ; 

with o as centre and any radius, cut OA, oa' in d, d'; 

draw df, gd' normal to ob, oc, and meeting each other in e ; 

through e draw normals to df, gd', and cut these normals, on 
their positive ends, by the circles fd, gd', i.e. by the cir- 
cles whose centres are f, g, and whose radii are fd, gd', 
in h, h'; 

join fh, gh'; 

then the plane angle hf-fe is equal to the diedral j3, 

and the plane angle eg-gh' is equal to the diedral y. 

For, revolve the right triangles feh, h'eg about fe, eg till 
eh, eh' are both normal to the plane a and coincide ; 
and revolve the right triangles dfo, ogd' about ob, oc till oa, 

oa' coincide in front of the plane a ; 
then*. *the right triangles feh, fed have coincident planes, 
the same base fe, and equal hypotenuse hf, df, 
.-.the perpendiculars eh, ed are equal. 

So, eh', ed' are equal, the points h,d, h',d' coincide, and 
the figure of lem.l is reproduced ; 
.'. the plane angles hf-fe, eg-gh' are equal to the die- 
drals ft, y. 



122 SPACE TRIGONOMETRY. [V, PR. 

To construct the diedral a, arrange the face angles in the 
order a, b, c or b, c, a, and then on as above. 

(b) Given the three diedrals, a, fi, y : 

Construct the polar triedral, taking angles equal to a, fi, y for 

the face angles ; 
then the three diedrals that are found are equal to the three 

face angles a> b, c that are sought. 

(c) Given a diedral and the two adjacent face angles, fi, c, a: 
Through any point o in the plane of the paper, draw rays oa, 

ob, -oc, making the angles aob, boc equal to c, a ; 
with o as centre and any radius, cut OA in D, and draw df 
normal to ob at f ; 



~~"~^.o 




through F draw a line such that the angle of df with this line 
is equal to the diedral fi, and cut this line on its nega- 
tive end by the circle fd, in h ; 

through h draw the normal to df at e ; 

draw eg normal to oc, cutting the circle od at d', and through 
d' draw oa'; 

then the angle coa is equal to the face angle b. 

The diedrals y, a may be constructed as in case (a). 

(d) Given a face angle and the ttvo adjacent diedrals, b,y, a : 

Construct the polar triedral, taking angles equal to b, y, a for 
a diedral and the two adjacent face angles ; 

then the face angle and two diedrals that are found are equal 
to the diedral and two face angles fi, c, a that are sought. 



1,§6.] GRAPHIC SOLUTION OF TRIEDRAL ANGLES. 123 

(e) Given two face angles and an opposite diedral, b, c, fi \ 
Through any point o in the plane of the paper, draw the rays 

oc, oa, ob, making the angles coa, aob equal to the face 

angles i, c ; 
with o as centre and any radius, cut oa in d ; 
through d draw gd normal to oc, and df normal to ob ; 
through F draw a line such that the angle df makes with this 

line is the angle fi, 
and with circle fd cut this line on its negative end in h ; 




through H draw eh normal to df, and with h as centre and 

radius gd cut de in k ; 
and through o draw oc ; , oc" tangent to the circle ek at g', g"; 
then either boc' or boc" is the face angle a ; and the other 
parts may be constructed as above. 
There is no triangle if gd<eh; one, a right triangle, if 
gd = eh; two, if hf>gd>eh; one, an isosceles triangle, if 
gd = hf ; one, if gd>hf. 

(/) Given two diedrals and an opposite face angle, /3, y, I : 
Construct the polar triangle, taking angles equal to /?, y, i for 

two sides and a diedral opposite one of them ; 
then the face angle and the two diedrals that are found are 

equal to the diedral and two face angles a, c } a, that are 

sought. 



124 



SPACE TRIGONOMETRY. 



[V, TH. 



§7. FOUR-PART FORMULAE. 

The reader will note the complete generality of the proof of 
the law of cosines and of the law of sines, no limitation of 
the sign or magnitude of any part being imposed, and the con- 
sequent generality of the formulae that depend upon these laws. 

THE LAW OF COSINES. 

Theor. 5. In a triedral angle : 

(a) The cosine of a face angle is equal to the product of the 
cosines of the other* two face angles less the product of their sines 
by the cosine of the opposite diedral : 
i. e. cos a — co si) cos c — sin i sin c cos a, 

cos i — cosc cos a — sin c sin a cos fi, 

cos c — cos a cos i — sin a sin I cos y. 




For, let o-ABC be a triedral angle, through D any point on the 
edge oa draw ed normal to the opposite face boc, 

and through ed, draw planes normal to the edges ob, oc, cut- 
ting ob in f, and oc in g ; 

then the lines df, fe are perpendicular to ob, and eg, gd to oc, 



5, §7. J FOUR-PART FORMULAE. 125 

and v the projections on oc of od and of the broken line ofed 
are equal, [df. 

and proj ed = 0, [ed perp. to oc. 

.'. proj od = proj of + proj fe, 
.-. proj od/od = proj of/od + proj fe/od, 
.\ proj od/od — proj of/of- of/od 

+ proj fe/fe • fe/fd • fd/od, 

.'. COS CO A = COS COB • COS BOA + COS OG-FE • COS FE-DF • Sill BOA, 

i. e. cos b — cos ( — a) • cos ( — c) + cos ( — a + r) • cos f3 • sin ( — c) ; 
.*. cos b = cos c cos a — sin c sin a cos /3. [II, theors. 5, 6. 

Q.E.D. 

So, '.'the projections on ob of od and the broken line oged 
are equal, 
.\ cos c = cos a cos b — sin a sin b cos y. q. e. d. 

So, if d be taken a point on oc, and ed be normal to the face c ; 
then cos a — cos b cos c — sin b sine cos a, q.e.d. 

The reader may well examine this proof with care : he will 
see that it is conclusive ; but he may ask what suggested the 
several steps in the tenth and eleventh lines. Only this: it was 
necessary to eliminate the lines which appear in the equation 

proj od == proj of + proj fe, 
and to bring in the ratios. 

Dividing by od, the first ratio proj od/od appears at once 
as one of the ratios sought. But proj of/od is not such a 
ratio, and the line of that joins the projection of of to od is 
used as an intermediary line ; then the ratio proj of/od is 
written as the product of the two ratios proj of/of, of/od, 
which can be interpreted. 

So, the ratio proj fe/od cannot be interpreted, and the 
two lines fe, fd that join the projection of fe to od are used 
as intermediary lines ; then the ratio proj fe/od is written 
as the product of the three ratios proj fe/fe, fe/fd, fd/od, 
which can be interpreted. 



126 SPACE TRIGONOMETRY. [V, TH. 

(b) The cosine of a diedral angle is equal to the product of the 
cosines of the other tivo diedrals less the product of their sines 
by the cosine of the opposite face angle : 
i. e. cos a = cos fi cos y — sin fi sin y cos a, 

cos fi — cos y cos a — sin y sin a cos b, 

cos y — cos a cos fi — sin a sin fi cos c. 

For, let a' , V ', c\ a', fi' , y' be the parts of a trieclral polar to 

the given triedral, 
then'.' a' = a, b' = /3, c' — y, a' = a, fi ' = b, y' — c, [theor. 4. 
and cos V — cos c f cos a' — sin c' sin a' cos /?', [above. 

.'. cos/?=:cos y cos a — sin y sin a cos b. 

So, cos y — cos a cos fi — sin a sin fi cos c, 

cos a = cos fi cos y — sin fi sin y cos a. Q. E. D. 

Cor. 1. cos^a— \/[sin (s — b) sin (s — c)/ sin b sine], 

cos\a— \f\_sin {a — fi) sin {a — y)/sinfi sin y], 

[e = ±(a + fi + y). 

For v 2cos 2 ^# = l + cos<* [II, theor.13, cor. 

= 1+ (cos b cos c — cos a) /sin b sin c [(«). 

= (cos b cos c + sin b sine — cos a)/sin b sin c 
= [cos (b -c) — cos «]/sin b sin c [II, theor. 11. 
= —2 sin -% (b — c + a) sin % (b — c — a) /sin b sin £ 

[II, theor. 12. 
— 2 sin | (a — b + c) sin -J- (« -f b — c)/sin b sin c 
= 2 sin (a — 5) sin (s — c)/sin 5 sin e, 
/. cos^a= V[sin (s — 5) sin (s — c)/sin 5 sine] q.e.d. 

So, v 2 cos 2 |a = 1 + cos a 

= 1 -r (cos /? cos y — cos a:) /sin fi sin ;/, 
, # . cos|a=: V[sin {a — fi) sm{6 — y)/sm fi sin;/]. 

Q.E.D. 



5, §7.] FOUR-PART FORM CLiE. 127 

Cor. 2. sin^a — s/[sin s sin (s — a)/sin b sin c], 
sin ia = *J\sin a sin (a — a)/ sin /3 sin y\ 

For v 2 sin a £ar = l — cos a [II, theor. 13, cor. 

= 1 — (cos b cos c — cos «)/sin b sin c 
= [cos a — cos (b -f e)]/sin b sin e [II, theor. 11. 
= — 2 sin i(a + b + c) sin ^ (« — b — <?)/sin J sin c 

[II, theor. 12. 
= 2 sin s sin (5 — a)/sin b sin c, 
.*. sin-^ — \/[sins sin (s — a)/sinb sine]. q.e.d. 

So, '.' 2 sin 2 ^a = l — cosa 

= 1 — (cos ft cos y — cos a) /sin ft sin ^, 
.'. sin^= V[sin (T sin (<7 — «)/sinyS sin y\ 

Cor. 3. tan^a— \J\sins sin (s — ct)/sin (s — b) sin (s — c)~\, 
tan ^a = V [sm (? sin{a — a) /sin (<r — ft) sin(a — >')]. 

Cor. 4. If a, b, c, a, ft, y be all positive and less than two 
right angles, and a, b, c be the interior diedrals, then : 

cos a — cosb cos c + sin b sin c cos a, 
cos k— — cos b cos c + sin b sin c cos a ; 

sin ^-A = V [sm (5 — J) sm (s — c)/sin b sin c], 
sin \a — V [sin e siw (a — ~E)/sin b s^ c] ; 

[e = ^-(a + b + c-2r). 

cos f A = V [s^ i ^iw (5 — a) /sin b sin c], 

cos \a — V [sitt (b — e) 5/ n (c — ~E)/sin b sm c] ; 

tan^K— \/[si?i (s — b) sin (s — c)/sins sin(s — a)], 
tan \a = V [si?& e sin (a — E)/sm (b — e) stw (c — e)] . 

For v A, B, c are supplementary to a, ft, y, 

.*. cos a — — cos A, cos/5=— cosb, cos y—— cose, 
sin # = sin a, sin ft — sin b, sin y — sin c, 

(jr = SUpE, 6— a — K — E, (T — ft = B — E, (T— ^ = C — E, 



128 



SPACE TRIGONOMETRY. 



[V, TH. 



THE LAW OF SINES. 

Theor. 6. In a triedral angle, the sines of the face angles are 
proportional to the sines of the opposite diedrals. 
i.e. sin a/ sin a = sin h/sin (3 = sin c/sin y. 

For let o-ABC be a triedral angle, through D any point on the 
edge oa draw ed normal to the opposite face bog, 

and through ed, draw planes normal to the edges OB, oc, cut- 
ting ob in F, and oc in g ; 




then the lines df, fe are perpendicular to ob, and eg, gd to oc ; 

and v the projections of the broken lines ofd, ogd on ed are 

equal, [df. proj. 

and proj of = 0, proj og = 0, [of, og perp. to ed. 

.'. proj FD = proj GD, 

.*. proj FD/oi) = proj gd/od, 

.*. proj fd/fd • fd/od = proj gd/gd • gd/od, 
i.e. sin FE-DF-sinOB-OD = sin EG-GD-sin oc-od ; 

.*. sin ( — (3) sin ( — c) — sin y sin 5, 

.'. sin /3 sin c — sin y sin h and. sin 5/sin (3 = sin c/sin y. 



6, §7.] FOUR-PART FORMULAE. 129 

So, if d be taken a point on oc, and ed be normal to the face c, 
then sin a/sin a = sin b/shi f5 ; 

.'. sin#/sintf = sin5/sin/? = sinc/sm;/. q.e.d. 

Cor. If a, b, c, a, fi, y be cell positive and less than two right 
angles, and A, b, c be the interior diedrals, then : 

sin a J sin A = sin b/sin B = sin c/sin a 
For v the angles**, A are supplementary, and so are /3,b and y,c, 
.'. sin A = sin a, sin b = sin/?, sine — sin y. [II, theor. 8. 

Note 1. If the theorem be regarded as relating to a spher- 
ical triangle, it may be written : The sines of the sides of a spher- 
ical triangle are proportional to the sines of the opposite angles; 
and the law of cosines may be expressed in like form. 

QUESTIONS. 

If abc be any spherical triangle, then : 

1. sin \a sin %/3/cos \y — sin s/sin c, 

2. sin \a sin \b /cos \c — sin <?/sin y, 

3. cos^-a- cos^/?/cos \y — sin (s — c)/sin<?, 

4. cos-|a cos^b /cos^c = sin (<7 — y)/smy, 

5. sin \a cos |/?/sin \y — sin (s — #)/sin c, 

6. sin -i- a cos I b /sin |c = cos (<r — a)/smy, 

7. cottar/cot \y — sin (5 — e)/sin (5 — a)> 

8. cot|#/cot|c = sin (& — y)/sm(<? — a), 

9. cot %/3 cotiy = — sin (s — a) /sin 5, 

10. cot £J cot ^ = — sin (<r — a)/sin a, 

11. sin(s-a) cot|a:=:sin(s-S) cot ■£■/? = sin (5 - c) cot^y, 

12. sin((T — a) cot|-a= sin ((?-/?) cot J5 = sin((r — y) cotjtf. 



130 SPACE TRIGONOMETRY. [V, THS. 

Note 2. The law of sines may be proved by aid of the law 
of cosines as follows : 

For v cos a — cos b cos c — sin b sin c cos a, 
. '. cos a = (cos b cos £ — cos a)/sin b sin c, 
. \ sin 2 **, = 1 — cos 2 **, = 1 — (cos b cos c — cos #) 2 /sin 2 b sin 2 £, 
.'. sin 2 <z/sin 2 « 

= [sin 2 # sin 2 6*— (cos b cos c — cos #) 2 ]/sin 2 # sin 2 Z> sin 2 £ 
= [1 — cos 2 # — cos 2 # — cos 2 c 

-f 2 cos a cos b cos c]/sin 2 a sin 2 # sin 2 c ; 
and v this value is symmetric as to a, b, c, 

.*. sin 2 /?/sin 2 #, sin 2 ;//sin 2 c have the same value, 
.*. sin^/sin 2 ^ = sin 2 /3/sin 2 # = sin 2 //sin 2 c. Q.e.d. 

Or as follows : 

V sin 2 a:=: [sin 2 5 sin 2 c— (cos b cos c — cos «) 2 ]/sin 2 # sm 2 6', 
= [sin b sin c — cos b cos c + cos a] 

• [sin b sin c + cos 5 cos c — cos #] /sin 2 Z> sin 2 c 
= [cos a — cos (b + c)] • [cos (& — c) — cos#]/sin 2 S sin 2 £ 
= 4 sin -J (# + 5 + c) • sin -J (5 + c — a) • sin \ (a — b + c) 

• sin ^ (# + & — c)/sin 2 S sin% 
.'. sin 2 a/sin 2 & = 4 sin 5 • sin (s — a) • sin (s — b)- sin (5 — c) 

/sin 2 # sin 2 # sin 2 £, 
and this value is symmetric as to a, b, c, 

.'. sin 2 6v/5in 2 « = sin 2 /3/sin 2 5 = sin 2 ^/sin 2 (;. q.e.d. 

Note 3. By v. Staudt the expression 

V (1 — cos 2 a — cos 2 5 — cos 2 c + 2 cos a cos b cos c) 

is called the sine of the spherical triangle, and the expression 

V (1 — cos 2 a — cos 2 /? — cos 2 / + 2 cos a: cos /? cos y) 

the S'me 0/ the polar triangle. Casey has called them the^rs^ 
staudtian and second staudtian. The first staudtian is equal 
to either of the products, 

sin b sin c sin a, sin c sin a sin /?, sin a sin b sin y ; 
and the second staudtian to either of the products, 

sin /? sin y sin a, sin y sin # sin b, sin a: sin yS sin c. 



6,7,§8.] ANGLES BETWEEN LINES IN SPACE, ASD PLANES. 131 



§8. ANGLES BETWEEN LINES IN SPACE, AND BETWEEN 

PLANES. 

Let ox, oy. oz be three lines through a point o, so directed that 
each is normal to the plane of the other two taken in 
order, 

i.e. so that ox is normal to the plane yoz, oy to zox, oz to xoy ; 

then also the angles yoz, zox, xoy are positive right angles as 
seen from x, y, z. 




Let op be any other line through o ; 

then op is completely determined by the angles xop, yop, zop. 

Let l 9 m, n = cos xop, cos yop, cos zop ; 

then I, m, n are the direction cosines of op, and determine it. 

So, I, m, n and a point determine a plane through the point, 
normal to op, and I, m, n are called the direction cosines 
of the plane. 

The direction cosines of a line not through o are the direc- 
tion cosines of a line through o parallel to the given line. 

Theor. 7. If I, m y n he the direction cosines of a line in space, 
then i 2 + m 2 + n 2 = l. 
For draw op parallel to the given line, through p draw a line 

parallel to oz, meeting the plane xoy in b ; 
through b draw a line parallel to oy, meeting ox in A ; 



132 



SPACE TRIGONOMETRY. 



[V, THS. 



thenv oa, ab, bp are the non-parallel edges of a rectangular 
parallelopiped and op is its diagonal,, 
.\ oa 2 + ob 2 + bp 2 = op 2 and oa 2 /op 2 + ob 2 /op 2 + bp 2 /op 9 = 1 ; 
i.e. P+m* + to*=l: q.e.d. [df . dir. cos. 

Theor. 8. If I, m, n, T , m', n' be the direction cosines of 
two directed lines in space, and a their angle, then : 
cos a — IV + mm' -4- nn '. 

For through o draw op, op' parallel to the two given lines, and 
draw bp normal to the plane xoy at B, and AB normal 

to ox at A ; 




project op and the broken line oabp on op'; 
thenv proj op = proj OA + proj AB + proj BP, 

.-. proj op/op = proj OA/op + proj AB/op + proj bp/op 

= proj oa/oa • oa/op -f- proj ab/ab • ab/op 

+ proj bp/bp • bp/op ; 
and v OA = proj op on ox, 

ab = proj op on oy, 

BPrrproj 0P on OZ, 

. \ cos a — cos xop' • cos xop 4- cos yop' • cos yop 

-f coszop'-coszop, 
i.e. cos a — 11' +mm f + nn f . q.e.d. 

Cor. If a be a right angle, then IT + mm' + nn r = 0. 



7-9, §8.] ANGLES BETWEEN LINES IN SPACE, AND PLANES. 133 

Theor. 9. If I, m, n, l\ m' , n' be the direction cosines of 
two directed lines that meet in space, a their angle, and X, jj, v 
the direction cosines of their plane, then : 
X = {mn' — m 'n) / sin a, 
fj. = {nV — n'l)/sin a, 
v — (lm' — I'm) /sin a. 
For '." the normal to the plane is perpendicular to every line of 
the plane, and so to the two given lines, 
.". lX+m/A-hnv = 0, l'X + m'p + n'v = Q ; [theor. 8, cor. 
and v X 2 + M 2 + ^ = h [theor. 7. 

,". A= {mn' — m'n) [solve for A. 

/ V [{mn' - m'n) 2 + (nV - n'l) 2 + {lm' - I'm)*] ; 
and v l 2 + m 2 + n 2 = 1, p + m" + n" = 1, [theor. 7. 

.*. X = {mn' — m'ri)/sj\l — {W + mm' + nn') 2 ], 

= {mn' — m'n) /sin a ; and so for jj, v. q.e.d. 

Note. A new proof of the law of cosines may be made 
from the principles established in theors. 7, 8, 9. 
Let a, fi, y be three directed lines in space that meet and 
form a triedral angle, whose face angles are a, b, c and 
whose opposite diedrals are a, /?, y, as shown in § 4 ; 
let I, m, n, T , m', n' , I", m", n" be the direction cosines of 
the lines a, J3, y, 

and X, ju, v, X' , jj.' , v\ X" , jj.", v" be the direction cosines 
of the planes fly, ya, a/3, i.e. of the planes a, b, c ; 
then*. A = {m'n" — m"n') /sin a, X' = {m"n — mn") /s\n b, [th.9. 
ju={n'l"~ n"l') /sin a, jj' — {n"l — nl")/s\n b, 
v = {Vm" - 1" my An a, v' = {l"m - lm") /sin b ; 
and v cosy =XX' + jjjj! + vv', [theor.8. 

/.cosy =[{7n'n"-m"n') {m"n-m?i")-\-{ril" -n"V) 

{n"l - nl") + {I'm" - l"m') {l"m - lm")]/sin a sin b 
= [{l'l" + m'm" + n'n") {l"l + m"m + n"?z) 

-{ll' + mm' + nn') (l ,n + m m '+n m )]/$ina sin b 
= [cos a cos b — cos c]/sin a sin b, [theors. 7, 8. 
.'. cos c — cos a cos b — sin a sin b cos y. 



134 SPACE TRIGONOMETRY. [V, THS. 

So, v cos a — X'X" + m'm" + yr " 

= (cos b cos c — cos tf)/sin & sin c, 
.'. cos a = cos b cos c — sin b sin c cos a\ 

So, '/cos/3 =\"\ + pi"/x+r"v 

= (cos c cos « — cos J)/sin c sin a, 
.\ cos b = cos 6* cos ft — sin c sin a cos /?. 

So, v cos c = IV 4- mm' 4- nn' 

= (cos a -f cos /? — cos ;/)/sin «r sin /?, 
.". cos ^ = cos a cos /? — sin a sin /? cos c ; 
and so for cos a, cos /?. 

§9. FIVE-PART FORMULAE. 
Theor. 10. In a triedral angle whose parts are a, b, c, a, /?, y, 

sin b cos y + cos c sin a 4- sin c cos a cos fi — 0, 
sin c cos /? 4- cos b sin a 4- sin b cos a cos y = Q ; 
sin c cos a + cos a sin b 4- sin a cos b cos y = 0, 
sin a cosy 4- cos c sm 5 -1- sin c cosb cos a = ; 
sm # cos fi-\-cosb sin c 4- stVz b cos c cos a — 0, 
sm b cos a-h cos a sin c + sin a cos c cos /? = ; 
For, project the broken line gofe on eg ; 
then'.* proj go = 0, [go perp. to ge, 

.'. proj of + proj FE = GE, 
.*. proj of/od 4- proj fe/od = ge/od, 
.'. proj of/of • of/od 4- proj fe/fe ■ fe/fd • fd/od 
= ge/gd ■ gd/od, 

.'. COS EG-OF • COS BOA + COS EG-FE • COS FE-FD • sill BOA 

= cos EG-GD • sin COA, 
i. e, cos ( — R — a) • cos ( — c) + cos ( — a) • cos (—/?)• sin ( — c) 

= cos }/-sin J ; 
V — sin « cos c — cos a cos /? sin c = cos y sin J, 
.'. sin b cos ^ 4- cos c sin # 4- sin c cos a cos /3 = 0. Q. e. d. 



9, 10, §9.] FIVE-PART FORMULAE. 

So, project the broken line foge on fe ; 



135 




E G 



thenv proj fo = 0, [fo perp. to fe. 

.*. proj OG + proj ge = fe, 
.\ proj OG/oD + proj ge/od = fe/od, 
,\ proj og/og • og/od + proj ge/ge • ge/gd • gd/od 
= fe/fd • fd/od, 
t. e. cos (r — a) • cos co a + cos fe-ge • cos ge-gd - sin coa 
= cos fe-fd • sin BOA ; 
.'. sin a -cos b + cos (2r + #) -cos ( — sup y) -sin 5 

= cos ( — /?) • sin ( — c), 
:. sin a cos b + cos a cos y sin J = — cos /? sin £, 
.*. sin c cos /? + cos /? sin # -f sin 5 cos a cos y = 0. Q. e. d. 
So, with normals drawn to the planes b, c, in turn, the other 

four formulae may be proved directly ; 
or they may be inferred by symmetry, from the two formulae 
just proved, i.e. the third and fifth from the first, and 
the fourth and sixth from the second. 



136 SPACE TRIGONOMETRY. [V, THS. 

Cor. sin ft cosy + cos c sin a -{-sin y cos a cos j3 = 0, 
sin y cos J3 + cos b sin a + sin /? cos a cos y — ; 
sin y cos a + cos a sin /? + sin a cos b cos y — 0, 
sin a cos y + cos c sin /? + sin y cos b cos a — Q ; 
sin a cos /? + cos b sin y + sin /? cos c cos a — 0, 
sin fi cos a + cos a sin y + sin a cos ccosft — 0. 
For v sin #/sin a — sin Z>/sin /? = sin c/sin y y [law of sines. 

.". sin a, sin b, sin c may be replaced by sin a, sin /?, sin y 
in the formulae of the theorem, and those of the corol- 
lary result directly. 

Note 1. The formulae of the theorem and those of the corol- 
lary may be paired in such manner that, if one of them be taken 
as applying to a triangle, the other is seen to be true for the 
polar triangle. 

E.g. the fourth formula of the corollary may be paired with the 
first formula of the theorem. 

Such a pair of formulas may be called a pair oi polar formulae. 

Note 2. The law of cosines may be proved by aid of the for- 
mulae given above, and without the polar triedral. 
E.g. Multiply the first formula of the corollary by cos/? and 
subtract the product from the last formula ; then cos c 
is eliminated ; 
and y sin/? cos # — sin/? cos/? cos }/ + cos # sin^(l — cos 2 /?) = 0, 
.*. cos a=cos ft cos y — sin/? sin y cos a. q.e.d. 

So, conversely, the formulae of theor. 10 may be found from 
the law of cosines by retracing these steps. 

QUESTIONS. 

Which formula of the corollary may be paired with the 
second formula of the theorem ? with the third ? with the 
fourth ? with the fifth ? with the sixth ? 

Eewrite the formulae of the corollary so as to show their cor- 
relation with those of the theorem, letter for letter and term 
for term. 



10,11, §9.] five-part formulae. 137 

napier's analogies. 
Theor. 11. In a triedral angle whose parts are a, b, c, a, /?, y, 
tan%(fi + y)/tan^a = -cos ^(b-c)/cos %(b + c), 
tan\{fi — y)/tan\a — — sin%(b-c)/sin%(b + c), 

tan %(b + c)/tan \a = - cos %{fi - y)/cos £(/? + y), 
tan %{b - c)/tan \a — - sin %(/3 - y)/sin i(/3 + y). 
For, add the fourth and fifth equations of theor. 10, 
then sin a (cos /3 + cos y) + (1 + cos a) sin {b + c) = ; 
and v sin a/sin a = sin 5/sin /? = sin e/sin y 

= (sin 5 + sin c)/(sin J3+ sin ?/), 
/. sin a — (sin 5 + sin c) sin ^/(sin /? + sin y), 
.'. (1 + cos a) sin (5 -f c) 

= — (sin 5 + sin c) sin # (cos /? + cos y)/(sin /? + sin ;/), 
.". (sin/? + sin^/)/(cos/?-}-cos y)- (1 + cos a:) /sin a: 
= — (sin 5 + sin c)/sin (# + c) ; 

.\ 2 sin |(/J + ;/) cos £(/?-;/) 

/2 cos £(/? -f ;/) cos J(/J — ;/) • cot ^a: 
= — 2sini(J + e) cos£(5 — c)/2sin £(# + c) cos^-(J + e), 

[II, theor. 12. 
.". tan %(/3-t-y)/t<dii±a= — cos ^(b — c) /cos ^(b + c). q.e.d. 
So, V sin#/sin # = (sin 5 — sin c)/(sin /? — sin ^), 
.". (l + cosa') sin (5 + e) 

= — (sin b — sin c) sin # (cos /> + cos y)/(sm j3 — sin y), 
and tan J(/? — j/)/tan Jar = — sin £(# — c)/sin J(J + c). 
So, add the first and second equations of theor. 10, cor., 
then sin a (cos b + cos 6') + (1 + cos a) sin (j3 + y) = ; 
and v sin a* = sin a (sin/? + sin;/)/(sin5 + sine), 
.\ (1 + cos a) sin (j3 + y) 

= — (sin /3 + sin )/) sin a (cos 5 4- cos c)/(sin b + sin #), 
and tanj(5 + c)/tan£a=: — cos J(/J — ^)/cos J(/J + ^). 



138 SPACE TRIGONOMETRY. [V, THS. 

So, v sin a=sin a (sin /? — sin ;/)/(sin J — sine), 
.-. (1-1- cos a) sin (/? + y) 

— — (sin /? — sin y)/sm a (cos (3 + cos ;/)/(sin 5 — sin c), 
and tan ^(b — e)/tan Ja = — sin ^(/? — ;/)/sin ^(/3 + }/). 

Cor. T/X b, c, a, /?, y be all positive and less than tivo right 
angles, and a, b, c be the interior diedrals, then : 

tan £(b + c)/cot ^-a = cos i(b- c)/cos }(b + c), 

tan £(b — c)/cot -J- a = si?i^(b — c)/sin ^(b + c), 

tarn J(i + c)/tan \a = eos £(b - c)/cos £(b -f c), 

taw £(& — 6) /tan \a — sin J(b — c)/s7?^(b -f c). 

Note. Another proof of Napier's analogies is given below : 
it does not use the formula of theor.10, and it has the single 
defect that it employs radicals, and so is not free from ambig- 
uous signs. 

tan \ (/? + y)/tnn \a [II, theor. 11, cor. 1. 

= (tan £/? + tan ^y)/tan \a (1 — tan £/? tan \y) 

/ sins sin (s — b) / sin s sin(s — c) 



+ 



* sin (s — c) sin (s — a) ' sin (s — a) sin {s — b) 



/ S1I16 1 i 

V sin (.<? — • 



/ sin 5 sin (s — a) [~ 
' sin (s — #) sin (s — c) L 



sin (5 — a)_ 

[theor. 5, cor. 3. 

Strike out the common factor Vsins, and multiply both 
numerator and denominator by Vsin (s — a) sin (s — b) sin (s — c); 
then tan ^{(3 + }/)/tan ^a 

— [sin (s — b) + sin (s — e)]/[s-in (5 — #) — sin 5] 

= sin |# cos J(i — e)/ — sin %a cos J( J + c) [II, th. 12. 

= — cos^(5-c)/cos£(& + e) ; q.e.d. 

and so for the rest. 



11, 12, §10.] SIX-PART FORMULAE. 130 

§ 10. SIX-PART FORMULAE. 

delambre's formulae. 
Theor. 12. In a triedral angle whose parts are a } b, c, a, /3, y, 

sin ht/sin ?a = =f sin\{b — c)/sin ^(fi — y), 
sin^a/cos ia= ±sini(b + c)/cos i(fi-y), 
cos ia/sin ±a = ± cos \(b — c)/sin %(/3 4- y), 
cos ia/cos \a~ =f cos i(b 4- c)/cos |(/3 4- y), 
with like formulae if a, a be replaced by b, j3 or by c, y: 
For V sin 2 ^/sin 2 ^=(l-f-cosa) (1 — cos«)/(l + cosar) (1 — cos a), 
and sin 2 tf/sin 2 tf = sin b sine/sin/? sin/, ' [law of sines. 

.*. (1 — cos a) /{I — cos a) 

= (1 4- cos a) sin b sin c/(l + cos a) sin /3 sin y 
= [(1 — cos a) — (1 + cos or) sin 5 sin c] 

/[(l — cos a) — (1 4- cos a) sin /? sin y\ [prop. 
= [1 — (cos a 4- sin b sin c cos or) — sin b sin c] 

/[l — (cos a 4- sin j3 sin ;/ cos a) — sin ft sin 7] 

= [1 — cos b cos c — sin 5 sin c\ 

/[l — cos /? cos 7 — sin fi sin ;/] [law of cos. 

= [1 — cos (b — c)]/[l — cos (/?— ;/)], [add. theor. 

.*. 2 sin 2 \a/% sin 2 \a—% sin 2 ^(i — c)/2 sin 2 ^-(/? — 7), 
.*. sin^a/sin^a= =Fsin^(5 — c)/sin J(/? — ;/). 
So, (1 — cos a) /{1 4- cos a-) 

= (1 — cos a) sin & sin c/(l 4- cos a) sin (3 sin y, 
and sin Ja/cos J# = ± sin £(& + #) cos^(/3 — y). 
So, (1 4- cos a) /{I — cos a-) 

= (1 4- cos a) sin 5 sin c/(l — cos a) sin /? sin ;/, 
and cos ^-a/sin £<* = ±rCos^-(5 — c)/sin^-(/?4-;K). 
So, (1 4- cos a) /(1 4- cos a) 

= (1 — cos a) sin b sin c/(l — cos a) sin /? sin y, 
and cos ^-a/cos 4-a: = +cos J(54-c)/cos4-(/5 + y). 



140 SPACE TRIGONOMETRY. [V, TH. 

Note 1. For any triangle the second members of these equa- 
tions must all have their upper signs or all their lower signs, 
since Napier's analogies may be got from Delambre's formulae 
by division, and must accord with them : 
first Nap. anal, from third Del. form, by fourth Del. form., 
second Nap. anal, from first Del. form, by second Del. form., 
third Nap. anal, from second Del. form, by fourth Del. form., 
fourth Nap. anal, from first Del. form, by third Del. form. 

Cor. If a, b,c, a, fi, y be all positive and less than tivo right 
angles, and a, b, c be the interior diedrals, then : 
sin \a/cos \k — + sin \{b — c) /sin ^(b — c), 
sin \a/sin \k — + sin %(b + c)/cos i(B — c), 
cos \a/cos -J-a = + cos \{b — c)/sin ^(b + c), 
cos \a/sin\K— -\-cos %(b + c)/cos |(b + c). 
For v A, B, c are the supplements of a, fi, y, 

.-. sin|^ = cos^-A- ■ •, sin^(/J — y)— — sin^(B — c)« • •_, 
and v the first members of these equations are positive, 
.'. the second members are position. 

Note 2. For any triangle, the equations 

sin %a/$m%a= q=sin ^(b — c) /sin ^(fi — y), 
sm%b/sin^fi = =psm J(c — #)/sin \{y — oi), ■ 
sin Jc/sin \y — =f sin \{a — S)/sin ^{a — fi), 
must be taken all with the upper sign or all with the lower 
sign ; and so of the other three groups of like equations. 
For •.".sin Ja : sin -Jar = — sinj(§ — c) : sin ^{fi—y) [up. signs. 
= sin^a — sin^(# — c) : sin^a + sin£(/J — y) 
= sm^a -f sin^-(5 — c) : sin \a — sin£(/? — y), 
.*. cos ^{s — c) sin^(s — S)/sin£(<r — y) cos %(<T — fi) 

= sin^(s — c) cos^{s — b)/cos^{(j—y) singer — fi), 
1] .*. tan j(s — b) cot ^-(5 — c) = cot ^(g — fi) tan £((7 — y). 
So, v sin^a : cos^a = + sin^-(5 + c) : cos ^(fi — y), [up. signs. 
2] .\ tan^-s cot ^(s — a) — cot^{a — fi) coti(a~y). 



12, §10. J SIX-PART FORMULA. 141 

So, vcos^a : sin£a= +cos£(ft — c) : sm%(/3+y) 3 [up. signs. 

3J .*. cot %(s — i) cot ^(s — c) — tan \a cot^a — a). 

So, *.• cos^a : cos|-^=: — cos^(b + c) : cos Hfi + y), [up. signs. 

4] .'.tanis tani(s — a) = coi^a cot 4-(<T — a). 

So, when the lower signs are taken : 
5] cot^(s-b) tan ^(s-c) = cot ±(&-/3) tan±((?-y), 
6] cot U tan±(s-a) = cot^(o~-/3) cot \(<5-y), 
T] taii\(s — ~b) tan^(,<? — c} = tan^-G'cot ^(<7 — <*), 
8] cot|-s cot \{s — ci) — cot\(j cot ^((5 — a) ; 

and with a, a replaced by i, (3, equations 5, 6, 7, 8 become : 
9] cot|-(s — a) tan ^(s — c) = coti(& — a) tan4-(<7— ;/), 

10] cot^s tan ^(s — h) = cot l(&- a) cot^(& — y), 

11] tan^-(s — a) tan ^-(s-c) = tanks' cot^-(<T — /?), 

12] cot ^-5 cot j(s — h) =cot|-(r cot^-((T — /?), 

with like formulae if a, # be replaced by c, y ; 

and v the products of the equations 2, 4, and of 10, 12, 
i.e. tan 2 ^s=cot%& cot\{a — a) cot^((T — j3) cot £((T — y), 
and cot 2 ^-s=:cot^(T cot^(<f— a) cot%(&—/$) cot^(G-y), 
are contradictory, and so of other pairs of equations, 

.*. the upper signs may not be used with the lower signs, 
i.e. the upper signs must be used together and the lower 

signs together ; 
and so for the other three groups of like equations ; 

.*. with the entire set of twelve equations, the upper signs 
must be used together and the lower signs together. 

Xote 3. Another proof of Delarnbre's formulas is as follows : 
For sin ^a • sin ^/3 = \l [sin s sin (s — a) /sin i sin c 

• sin s sin (5 — 5)/sin c sin a] 
— ± sin s/sin c • V [sin (5 — a) sin (s - V) /sin a sin 5] 
= ± sin s/sin c • cos \y. 



142 SPACE TRIGONOMETRY. [V, THS. 

So, si a \a cos i/3= ± sin (5 — a) /sin c • sin %y, 
cos^a sin %/3= ±sin (s — fy/smc'sm^y, 
cos \a cos ^(3 — ± sin (.5 — e)/sin c • cos |y. 

In these equations the upper signs go together, and the lower 
signs go together. 

For, multiply together the first two equations, 
then sin 2 ^o:-sin/?= ±sins sin (s — &)/sin 2 c-sin y 
= ± sin 2 \a • sin i sin c sin //sin 2 6*, 
.*. sin /? = d=sin S sin //sin 6*, 
.\ sin /?/sin h = ± sin //sin c ; 
and this equation is true only when the sign + is used, 
i.e. when the two upper signs are taken in the first two equa- 
tions, or the two lower signs ; 
and so of the other equations. 
Add the first and fourth equations, 
then cos^(a — /?) = [sin s + sin (s — c)]/sinc-cos^y 

= sin ^(a + 5)/sin \c • cos \y ; 
and so for the other formulae. 

Note 4. Simon FHuillier's formulas. If equations 2, 3 of 
note 2 be multiplied together, the products give the equation 
tan \s cot^(s — a) cot %(s — i) cot ^(s — c) 

= tan^(T cot%(<r—a) cot %(<? — /?) cot^(<7 — /) ; 
and, by substitution from other equations of the set, 

= cot%((T — f3) cot^((T — y) cot^(s — i) cot %(s — c), 
= cot J((X — /) cot|-(<7 — a) cot ^(,9 — c) cot J(s — a), 
= cot^-((T — a r ) cot J(<7 — /?) cot^(s — #) cot£(s — #), 
= tan^(7 cot^((? — a) tan^-s cot %(s — a), 
= tanker cot \(p — fi) tan ^-5 cot^(s — b), 
= tanker cot K^ — K) ^ an i s cot£(s — c), 
.*. tan ^-(T tan ^s = cot ^((? — a) cot J(s — a), 
= cot ^(6 — (3) cot £(s — 5), 

= COt^((7 — /) Cot£(s — tf). 



12, 13, §11.] 



THE RIGHT TRIEDRAL. 



143 



§11. THE RIGHT TRIEDRAL. 

Theor. 13. Inatriedralangle whose parts are a, b, c, a, ft, y, 
if y be a right diedral, then : 

sin a — sin c si?i a— — tan b cot ft, 
sin b — sinc sin ft = — tan a cot a, 
cos c —cos a cos b —cot a cot ft, 
cos a— — cos a sin ft = — tan b cot c, 
cos ft — —cosb sin a— —tana cote. 
For v y is a right diedral, 

.-. siny — ^), cos y = ; 
and v sin a/sin a — sin 5/sin ft — sin c/sin y, [theor. 6. 

.*. sin a /sin x — sin c, 
and sin 5/sin ft — sin c. 

So, v cos c = cos a cos b — sin a sin b cos y, [theor. 5. 

.*, cos c — cos a cos b. 
So, Y cos a = cos ft cos y — shift siny cos a, [theor. 5. 

.". cos a— — sin/? cos a. 
So, Y cos/? — cos/ cos tf — sin y sin a: cos#, [theor. 5. 

.*. cos /? = — sin a cos 5. 
So, Y cos )/ = cos # cos /? — sin a sin /? cos £, [theor. 5. 

.'. cos c = cot a cot /?. 
The four formulae below come from the six proved above, 
cos a — — sin ft cos a— — sin b/sin c • cos c/cos 5 = — tan b cot c, 
cos/? = — sin or cos b — — sin #/sinc« cos c/cos # = — tan # cote, 
sin a = sin c sin or = — sin b/sin ft • cos ft /cos b — — tan b cot /?, 
sin b — sin c sin ft — — sin a/sin a • cos ^/cos a— — tan a cot a\ 

In the figures below, the great circle a may stand for the 
earth's equator, as it would be if the earth revolved from east 
to west half the time ; the letters i\ T , s stand for the north and 
south poles, the great circle b for a meridian, and the great 
circle c for any other great circle cutting the equator and the 
meridian. The angle y is always a positive right angle. The 



1U 



SPACE TRIGONOMETRY. 



[V, TH. 



triangles appear as seen from the sun when fifteen degrees 
north of the equator. The invisible parts of the arcs are 
shown by the broken lines. 




Cor. If a, b, c, a, (3, y be all positive and less than tivo right 
angles, y a right diedral, and a, b, c the interior diedrals ; 
then sin a — sin c sin a = tan b cot b, 

sin b — sin c sin b ~ tan a cot a, 

cos c — cos a cos b — cot a cot b, 

cos a = cos a sin b = tan b cot c, 

cos B — cosb sin A = tan a cote. 



13, §11.] THE RIGHT TRIEDRAL. 145 

NAPIER'S RULES. 

By an ingenious device of Lord Napier these ten formulae 
are remembered by two simple rules : 

Ignore the right angle ; take the two sides, and replace the hy- 
potenuse and two oblique angles by their complements ; 

of the five parts so found call any one the middle part, the two 
lying next it adjacent parts, and theothers opposite parts; 

then : sni rmd-part — prod taw adja parts, 
sin nud-part = prod cos oppo parts. 

If the three parts considered lie together, that which lies 
between the other two is mid-part and the other two are adja- 
cent parts ; if two lie together and the third apart from them, 
the third one is mid-part and the other two are opposite parts. 




E.g. let o-ABC be an ideal triedral right angled at c ; 

then, of the three parts a, b, co-c, co-c is mid-part, a, b are 

opposite parts, 
and cos c = cos a cos b. 
So, of the parts co-a, co-b, co-c, eo-c is mid-part, co-A, co-B 

are adjacent parts, 
and cos c— cot A cot B. 

So, of the three parts co-a, co-B, a, co-A is mid-part, co-B, a 

are opposite parts, 
and cos A — sin b cos a. 



146 SPACE TRIGONOMETRY. [V, THS. 

§12. THE IDEAL TRIEDRAL. 

A triedral whose face angles and diedrals are all positive and 
less than two right angles is an ideal triedral. Two parts of 
such a triedral are of the same species if they be both acute, 
both right, or both obtuse. 

It has been shown in geometry that in an ideal triedral : 

1. The sum of the three face angles lies between naught and 
four right angles. 

2. The sum of the three interior diedrals lies between two 
right angles and six right angles. 

3. Each face angle is less than the sum of the other tw T o face 
angles, and so of the exterior diedrals. 

4. Each interior diedral is greater than the difference be- 
tween two right angles and the sum of the other two interior 
diedrals. 

5. Of two unequal face angles the greater lies opposite the 
greater interior diedral, and so opposite the less exterior die- 
dral, and conversely. 

6. If two face angles be equal, so are the opposite diedrals 
(both exterior and interior), and conversely. 

7. A plane through the vertical edge of an isosceles triangle 
perpendicular to the opposite face, bisects the interior vertical 
diedral and the opposite face angle. 

Certain other facts relating to ideal triedrals are manifest, 
and still others appear by examining formulae already proved. 

8. The sine of every part is positive. 

9. Every half part is positive and acute, and all its ratios are 
positive. 

10. The half sum of two parts is positive and less than two 
right angles. 

11. The half difference of two parts is acute and its cosine 
is positive. 



14, 15, §12.] THE IDEAL TRIEDRAL. 147 

THE IDEAL TRIANGLE. 

Theor. 14. In an ideal triangle, that one of two unequal 
sides which is nearer right lies opposite the angle which is nearer 
right, and conversely. 

For v that angle which lies nearest to a right angle has the 

greatest sine, 
and Y sin a/sin A = sin S/sin b = sin c/sin c, 

/.if sin a > sin b, then also sin A > sin B; and so of 
the others. 

Theor. 15. In an ideal triangle, a side and its opposite in- 
terior angle are of the same species, if another side he as near 
right as the given side, or if another angle be as near right as 
the given angle. 

For let the side c be as near right as the given side a ; 

then*. * cos c^ cos a, and cos5<l, 

.". cos b cosc< cos a, or cos b cosc = cos*a:=0, 
.'. cos a, cos a — cos J cos c, are positive, negative, or zero 
together ; 

and Y cos a = (cos a — cos b cos c)/sin b sin c, 

and sin b, sin c are positive, 

.*. cos#, cos A are positive, negative, or zero together, 
.\ a, A are both acute, both obtuse, or both right, q.e.d. 

So, let the angle c be as near right as the angle a ; 
then*.* cos a = (cos A + cos B cos c)/sin b sin c, 

and cos a, cos a + cos b cose are positive, negative, or zero 
together, [as above. 

.*. cosfl, cos A are positive, negative, or zero together, 
,\ a, A are both acute, both obtuse, or both right, q.e.d. 

Cor. A side and the opposite exterior angle may be both right ; 
but if one of them be acute the other is obtuse, if another side be 
as near right as a given side, or another angle be as near right 
as a given angle. 



148 SPACE TRIGONOMETRY. [V, THS. 

Theor. 16. In an ideal triangle the half-sum of tivo sides 
and the half -sum of their opposite interior angles are of the 
same species. 

For v cos \ (b -f c)/cos J (b + c) = cos Ja/sin Ja, 

and cos 2^, sin^A are both positive, 

.*. cos J (b + c), cos J (b + c) are positive, negative, or 
zero together, 

.\ ^{b + c), ^-(b + c) are both acute, both obtuse, or 
both right. q.e.d. 

Cor. The half-sum. of two sides and the half-sum of their 
tioo opposite exterior angles may be both right ; but if one of 
them be acute the other is obtuse. 

§ 13. IDEAL RIGHT TRIANGLES. 

A triangle having two right angles and two right sides is a 
biquadrantal triangle. 

Theor. 17. In an ideal right triangle, if another part be- 
sides the right angle be right the triangle is biquadrantal. 

For let c be the right angle in the triangle abc ; 
then*.* cos c— cos a cos b, 
cos a = cos a sin b, 
cos b = cos b sin a, 
and sin a, sin b =£ 0, [a, b are not zero nor two right angles. 

.". if a be right, then cos a = 0, 

.-. cos c, cos a = 0, and c, A are both right, q.e.d. 
So, if b be right, then c, B are right. q.e.d. 

So, if A be right, then cos A — 0, 

.\ cos a, cosc = 0, and a, c are both right, q.e.d. 
So, if b be right, then b, c are right. q.e.d. 

So, if c be right, then cos c — 0, 

.*. either cos a = 0, or cos 5 = 0, 

.'. a or b is right, and a or b is right. q.e.d. 



16-21, §13.] IDEAL RIGHT TRIAXGLES. 149 

Theor. 18. In an ideal right triangle, not biquadrantal, 
the hypotenuse is nearer right than either oblique side. 

For let c be the right angle in the triangle abc ; 
then*. • cos c — cos a cos b, cos a, cos£<l, 

.'. cose < cos a, cose < cos b, 

.". c is nearer right than a or b. 

Theor. 19. In an ideal right triangle, not biquadrantal, 
an oblique angle is nearer right than its opposite side. 

For let c be the right angle in the triangle abc ; 

their.- cos a = cos a sin b, and sin b < 1, [b not right. 

.\ cos a < cos a, 

.'. A is nearer right than a. q.e.d. 

Theor. 20. In an ideal right triangle, an oblique angle and 
its opposite side are of the same species. 

For let c be the right angle in the triangle abc ; 

thenv cos A = cos a sin b, and sin B is positive, 

.". cos A, cos a are positive, negative, or zero together, 
.". a, a are both acute, both obtuse, or both right, q.e.d. 

Theor. 21. In an ideal right triangle, if the hypotenuse 
be acute the ttvo oblique sides are of the same species, and so are 
the two oblique angles ; but they are of opposite species if the 
hypotenuse be obtuse. 

For let c be the right angle in the triangle abc ; 
thenv cos c — cos a cos b~ cot a cot b, 
and cos c is positive if c be acute, 

.'. cos a, cosb are both positive or both negative, and so 
are cot A, cot b ; 

.\ a, b are both acute or both obtuse, and so are A, b. 

So, V cose is negative if c be obtuse, 

.'. cos a, cos b are of opposite signs, and so are cot A, cot b ; 
/. a, b are of opposite species, and so are a, b. 



150 SPACE TRIGONOMETRY. [V, PR. 

QUESTIONS. 

1. sin 2 \c — sin 2 \a cos 2 \b + cos 2 \a sin 2 \b. 

2. tan 2 ^a = tan %(c + b) tan£(c — b). 

3. tan 2 |-A = sin (c — #)/sin (c + b). 

4. tan ^a = sin (c — S)/sin a cos b = sin # cos 5/sin (c + b). 

5. sin {a — b)~ sin a ten ^-A — sin b tan ^b. 

6. tan 2 ^# == tan £(b -f a — R)/tan ^(b — a + r). [r a rt. ang. 

7. tan 2 jk?= — cos (a + b)/cos(a — b). 

§14. SOLUTION OF IDEAL RIGHT TRIANGLES. 

PROB. 2. TO SOLVE AN IDEAL RIGHT TRIANGLE, GIVEN TWO 
PARTS BESIDES THE RIGHT ANGLE. 

Out of the formulce o/theor. 13, cor. select those which involve 
the two given parts and one of the parts sought, and solve 
these three equations for the three parts sought. 

Check : Substitute the three computed parts in that formula 
which involves them all, and see if they give an identity. 

Or, better, following Napier's rules : 

Take each of the two given parts in turn for middle part, and 
apply that formula which brings in the other given part; 

take the remaining part for middle part, and apply that for- 
mula which brings in both of the parts just found ; 

solve the three equations so found for the parts sought. 

Check : Make the part last found the middle part, and apply 
that formula ivhich brings in both the given parts. 

The check is defective in this, that it tests the logarithms, 
but not the angles got from these logarithms, i.e. not the final 
results : more perfect checks are got from any of the general 
formulae which involve the three computed parts and one or 
more of the given parts. 

The check is applied to the sine of the part last found ; if the 
two values got for this sine, natural or logarithmic, differ by 
not more than three units in the last decimal place, the work 



2, §14.] SOLUTION OF IDEAL RIGHT TRIANGLES. 151 

is probably right, since the defects of the tables permit this dis- 
crepancy in the two results : if such discrepancy exist, the mean 
of the two values may be used. 

These rules involve only the interior angles : for the general 
solution of the right triangle the formulas of theor. 13 are 
available. 

There are six cases. 

(a) Given a, b, the two sides about the right angle c : 
then*.* sin a = tan b cotB, .*. cot b = sin a cot b, 

sin # = tan a cot A, .*. cot A = sin b cot a, 

cos c = cot a cot b ; check: cos c = co$a cos b. 

One triangle is always possible and but one : for the species 
of A, B, c is shown by the algebraic signs of the cotangents or 
cosines that are used, or by theors. 20, 21. 

Geometrically, With the given arcs a, b at right angles, their 
extremities can be joined by a positive great arc, always in one 
way and in but one way. 

(b) Given c, a, the hypotenuse and one side : 

then cos c — cos a cos b, ,\ cos 5 = cosc/cos a, 

sin a — sin c sin a, .\ sin A = sin a/sin c 9 

cos b = cos b sin A ; check: cos B = tan^ cot c. 

A triangle is possible only when c is nearer right than a, or 
when c, a are both right. 

The species of b, b is shown by the sign of cos b, cos b, and 
A is of the same species as a. 

If c, a be both right, then cos 5, cos b are indeterminate, b, 
b are equal, and a is right. 

Geometrically. On a directed great circle b take any point 
c ; through c draw a great circle a perpendicular to b, and 
take B a point on a such that bc is positive and equal to the 
given arc a ; with b as pole, and an arc-radius equal to the given 
arc c, draw a small circle cutting the great circle b in two 
points ; take a one of these points such that ca is positive : 



152 SPACE TRIGONOMETRY. [V, PR. 

then the triangle abc is the triangle sought, and there is but 
one such triangle. 

If the arc c be not nearer right than the arc a, the small 
circle is wholly within or wholly without the great circle b and 
there is no triangle. 

(c') Given a, b, an oblique angle and the adjacent side : 
then sin b = tan a cot A, .*. tan a — sin b tan a, 

cos A = tan b cote, .\ cot c =cosa cot b, 

cos b = tan a cot c ; check : cos b — cos b sin a. 

One triangle is always possible and but one. The species of 
the computed parts are shown by the signs of the tangent, co- 
tangent, and cosine that are used, or by theors. 20, 21. 

Geometrically. On a directed great circle b, take c, A two 
points such that the arc ca is positive and equal to the given 
arc B ; at c draw the directed great circle a, perpendicular to 
the circle 5, and at a draw the directed great circle c making 
an angle with the circled equal to a, the supplement of the 
given angle A, and meeting the great circle a in two points ; 
take b one of these points such that bc, ca are positive arcs : 
then the triangle abc is the triangle sought, and there is but 
one triangle. 

(d) Given b, J, an oblique and its opposite side : 
then cos b = cos b sin a, .*. sin A = cos b/cos b, 

sin b — sin c sin b, .*. sin c — sin b/sin b, 

sin a — sin a sin c ; check : sin a — tan b cot B. 

If B, b be not of the same species, no triangle is possible, for 
then sin A is negative. 

If b be nearer right than B, no triangle is possible, for then 
sin a >1, which is impossible. 

If b, b be equal but not right, the triangle is biquadrantal, 
for then sin a, sin a, sine are all 1, and A, a, c are all right. 

If b, 5 be both right, the triangle is biquadrantal, for then 
c also is right, and a, a are indeterminate. 



2, §14. J SOLUTION OF IDEAL KIGHT TRIANGLES. 153 

If B be nearer right than b and of the same species, there 
are two triangles. 

For v A, a, c are all found from their sines, 
and sin a, sin a, sin c are all positive, 

.'. to each of these sines correspond two possible angles, 

supplementary to each other, both positive and less 

than two right angles. 

But a, a must be of the same species ; 
and if c be acute, A, a are of the same species with b, b. 
So, if c be obtuse, a, a are of species opposite to that of b, b, 
.'. two triangles, and but two, are possible. 

Geometrically. Let bdb', b'eb be half circles forming lines 
whose angle is the given angle b of the triangle ; draw an arc 
ca normal to bcb' and equal to the given arc b ; with its initial 
point c sliding over bcb' push the arc-normal ca to the right 
and left till the terminal point A rests on the circle b'eb ; 
then if b be acute and b<B, two triangles a'bc', a"bc"; 

if 5 = b, one triangle, biquadrantal ; if £>b, no triangle. 
So, if b be obtuse and 5>b, two triangles are formed ; 

if b — B, one triangle, biquadrantal ; if #<b, no triangle. 
(e) Given c, a, the hypotenuse and an oblique angle : 
then cos c = cot a cot b, .*. cot b = cos c tan a, 

cos A = tan b cot c, .*. tan b — tan c cos A, 

sin a = tan b cot b ; check: sin a = sine sin a. 
One triangle is always possible and but one, for the species 
of b y B is shown by the sign of the tangent and cotangent, and 
a, A are of the same species. 

Geometrically. Through a point a on a great circle b, draw 
the great circle c, making with the great circle b the given angle 
A ; lay off ab positive and equal to the given arc c ; from b 
draw the great circle a perpendicular to the circle b and meet- 
ing it in c ; then is the triangle abc the triangle sought, and 
this triangle can be drawn in but one way. 



154 SPACE TRIGONOMETRY. [V, PR. 

(/) Given A, b the two oblique angles : 
then cosa = cos# sinB, .*. cos a = cos A/sin B, 

cosb=tcos b sin A, .*. cos # = cosB/shiA, 

cos c — cos a cos b ; c7j0c& : cos c = cot a • cot b. 

The species of the parts sought is shown by the signs of the 
cosines : but the solution is possible only when cos A< sin B 
and cosb< sin a, 
i.e. when a is nearer right than co-B, and b than co-A. 

Geometrically. Let ad, a'e be great circles whose angle is 
the given angle a, and draw great arcs cb, c'b', c"b", • • •, nor- 
mal to the arc ada' ; then the angles cba vary from r — a to 
R + A, and one of them is the given angle b, if b be nearer right 
than co-A, and but one. 

QUADRANTAL TRIANGLES. 

Find the polar of the given triangle : it is a right triangle ; 
solve this triangle and take the supplements of the parts thus 
found for the parts sought in the given triangle. 

A biquadrantal triangle is indeterminate unless either the 
base or the vertical angle be given. 

ISOSCELES TRIANGLES. 

Draw an arc from the vertex to the middle of the base, thereby 
dividing the given triangle into two equal right triangles; 
solve one of these triangles. 

If only the base and the vertical angle be given, there are 
two triangles, one triangle, or none, according as the base is 
less than, equal to, or greater than the vertical angle ; if only 
the two equal sides or the two equal angles be given, there is 
an infinite number of triangles ; otherwise, subject to the con- 
ditions just found (b,f), there is one triangle, and but one. 

OBLIQUE TRIANGLES. 

In most cases a perpendicular may fall from a vertex of an 
oblique triangle to the opposite side in such manner that one 



2, §14.] SOLUTION OF IDEAL RIGHT TRIANGLES. 155 

of the right triangles thus formed contains two of the three 
known parts, and the other right triangle contains one of them. 

Solve these right triangles in order and so combine the parts as 
to find the parts sought in the given triangle. 

The solution of right triangles may take this form : 

Given a = 72°, b = 125°, c = 90°, then : 

cot A = cot a sin b, cot B=sin a cot b, 

9.511776 9.978206 

9.913365 9.845227 

9.425141 9.823433 neg. 

A = 75° 5' 45". B = 123° 39' 40".* [theor. 20. 

cos c = cot a cot b, check : cos c — cos a cos b. 
9.425141 9.489982 

9.823433 9.758591 

9.248574neg. 9.248573 neg. 

c = 100°12'34". [theor. 21. 

QUESTIONS. 

Solve these right triangles, given c, a right angle, and : 

1. a, 116°; b, 16°. [97° 39' 24", 17° 41' 40", 114° 55' 20". 

2. c, 140°; a, 20°. [32° 8' 48", 115° 42' 24", 144° 36' 29". 

3. A, 80° 10'; b, 155°. [67° 6' 23", 153° 57' 34", 110° 46' 40". 

4. a, 100°; a, 112°. [27° 36' 59", 109° 41' 49", 25° 52' 33", 

or 152° 23' 1", 70° 18' 11", 154° 7' 27". 

5. e, 120°; A, 120°. [131° 24' 34", 40° 53' 36", 49° 6' 24". 

6. A, 60° 47'; b, 57° 16'. [54° 31' 52", 51°43'1", 68° 55' 50". 

7. c, 140°; a, 140°. 8. c, 120°; a, 90°. 
Solve these quadrantal triangles, given : 

9. A, 80°; a, 90°; b, 37°. 10. b, 50°; b, 130°; c, 90°. 
Solve these isosceles triangles, given : 

11. a, 70°; b, 70°; A, 30°. [157° 39' 34", 134° 24' 30". 

12. a, 30°; a, 70°; b, 70°. 

13. a y 119°; 5, 119°; c, 85°. [113° 57' 11", 72° 26' 22". 



156 SPACE TRIGONOMETRY. [V, PR. 

§15. SOLUTION" OF IDEAL OBLIQUE TRIANGLES. 

PROB. 3. TO SOLVE AN IDEAL OBLIQUE TRIANGLE, GIVEN 
ANY THREE PARTS. 

Apply such of the formula of theor.b cor A, theor.6 cor., theor. 
11 cor., as serve to express the three parts sought in terms 
of known parts. 

Where possible the computer will choose his formulae so as to 
avoid angles near the ends of a quarter. 

Check -.form an equation that involves the three computed parts 
and such of the given parts as may be necessary. 
If the equation so formed be a true equation the parts have 
probably been computed correctly. - 

DelarnbiVs formulae are useful as checks, and so are the for- 
mulae shown in the questions in § 13. 

In the check the parts must be involved by different ratios, 
or in different combinations, from those used in the solution. 

Note. These rules involve only the interior angles : for the 
general solution of the oblique triangle, the formulae of theors. 
5, 6, 11 and their corollaries are available. 

There are six cases : 

(a) Given b, c, a, two sides and the included angle : 
then tan £(b + c) = cot Ja • cos %(b — c)/cos J(i + c), 
tan £(b — c) — cot J A • sin \{b — c)/sin l(b -f- c), 

B = £(B + 0) +1(B - C), C = ±(B + C) - i(B - C), 

tan \a — tan %(b + c) • cos J(b + c)/cos £(b — c). 
Chech : the law of sines or one of Delambre's formulae. 

There is always one triangle and but one. 
For v whatever the values of b, c, A, the parts given, 

tan|-(B + c), tan£(B — c), tan^a are always possible, 
and the species of J(b -I- c), J(b - c), \a are shown by the 
signs of their tangents, 
.-. b, c, a are always possible, and they have single values. 



3, §15. J SOLUTION OF IDEAL OBLIQUE TRIANGLES. 157 

Geometrically. Lay off the arc ca equal to the given arc b; 
at A turn by the angle a, the supplement of A, and lay off the 
arc ab equal to the given arc c ; join bc : the triangle ABC is 
the triangle sought, and with the data there is always one and 
but one such triangle. 

(b) Given b, c, a, two angles and their common side: 
then tan ^{b -\-c) = tan \a • cos J(b - c)/cos |(b -f c), 

tan \(b — c) — tan \a • sin J(b — c)/sin \ (b 4- c), 
S=i(* + c)+K*-c) J c=t(b+c)-}(b-c), 

cot |-a = tan J(b + c) • cos J(5 -f c)/cos ^{b — c) . 
Check: the law of sines, or one of DelambiVs formulae. 

There is always one triangle and but one. 
For v whatever the values of b, c, a, the parts given, 

tan \{l) + c), tan ^(b — c), tan Ja are always possible, 
and the species of %(b + c), i{b — c), -J A are shown by 
the sign of their tangents, 
.*. 5, c, A are always possible, and they have single values. 

Geometrically, At any point b, on an indefinite arc ab, turn 
by the angle ft, the supplement of b, and lay off the arc bc equal 
to the given arc a ; at c turn by the angle y, the supplement 
of c, and draw an arc cutting ab in A : the triangle abc is the 
triangle sought, and with the data there is always one and but 
one such triangle. 

Note. This triangle may be solved under case (a), using the 
polar triangle whose parts V , c ! , a' are supplementary to b, c, a> 
and the computed parts b', c', a', to b, c, a, the parts sought. 

(c) Given b, c, b, tivo sides and an angle opposite one of them : 
then sin c = sine- sin B/sinZ>, 

cot Ja = tan J (b + c) cos \ (b + c)/cos \ (b — c), 
tan^ = tan^(£-f c) cos \ (b + c)/cos J (b — c). 
Check : one of Delambre's formulae. 

If b, c, B be all right, the triangle is biquadrantal, and A, a 
are indeterminate and equal. 



158 SPACE TRIGONOMETRY. [V, PR. 

If sin c sin b > sin b, then sin c > 1, which is impossible, 
and there is no triangle. 

If sin c sin b = sin b, then sin c = 1, c is right, and there 
is one (a right) triangle if b, b be of the same species, but no 
triangle if they be of opposite species. 

If sin c sin B < sin b, then sin c < 1, and c may be either 
of two supplementary angles ; but these angles must be taken 
subject to the law, the greater angle lies opposite the greater side. 

In particular : 

If c be nearer right than b, there are two triangles if b, b be 
of the same species, but none if they be of opposite species. 

If c be just as near right as b, there is one (an isosceles) tri- 
angle if b, b be of the same species, but no triangle if they be 
of opposite species ; and c, c are also of the same species. 

If c be less near right than 5, there is one triangle and c, c 
are of the same species. 

Geometrically. Lay off an arc ab equal to the given side c ; 
at b turn by the angle /?, the supplement of b, and lay off the 
indefinite arc bc ; with A as pole and an arc-radius equal to b 
describe a small circle : 
if this small circle neither cuts nor touches the circle BC, there 

is no triangle ; 
if it touches the arc bc at a point c, such that bc is positive 
and less than two right angles, there is a right triangle ; 
if it cuts the arc bc at one point c, such that bc is a limited 

arc, there is one triangle ; 
if it cuts the arc bc in two points c 19 c 2 , such that Bc a , bc 2 are 
both limited arcs, there are two triangles. 

(d) Given b, c, b, two angles and a side opposite one of them : 
then sin c = sin c • sin J/sin b, 

tan \a — tan j(b + c) cos J (b + c)/cos \ (b — c), 
cot jA = tan^ (b + c) cos^- (5 + c)/cos \ (b — c). 
Chech : one of Delambre's formulae. 

If b, c, b be all right, the triangle is biquadrantal, and a, A 
are indeterminate and equal. 



3, § 15.] SOLUTION OF IDEAL OBLIQUE TRIANGLES. 150 

If sin c sin b > sin b then sin c > 1, which is impossible, 
and there is no triangle. 

If sine sin b = sin b, then sin£ = l, c is right and there 
is one (a quadrantal) triangle if b, b be of the same species, 
but no triangle if they be of opposite species. 

If sin c sin b < sin b, then sin c < 1, and c may be either 
of two supplementary arcs ; but these arcs must be taken sub- 
ject to the law that in an ideal splierical triangle the greater 
angle lies opposite the greater side. In particular : 

If c be nearer right than B, there are two triangles if b, b be 
of the same species, but none if they be of opposite species. 

If c be just as near right as b, there is one (an isosceles) tri- 
angle if b, b be of the same species, but none if they be of op- 
posite species. In this triangle c, c are also of the same species. 

If c be less near right than b, there is one triangle, and c, c 
are of the same species. 

Geometrically. Draw an indefinite arc ; at any point B, turn 
by the angle f3, the supplement of B, and draw the indefinite 
arc bc ; at any point c, turn by the angle y, the supplement of 
c, and draw the arc ca equal to b ; let the arc first drawn slide 
along the arc bc, as aline may slide along one of its bounding 
circles, without changing the angle b. 

If this sliding arc do not pass through the point a, there is 
no triangle ; if it touch A, and the angle a be a limited angle, 
there is one (a quadrantal) triangle ; if it pass through A twice, 
so as to make the two angles A v a 2 both limited angles, there 
are two triangles. 

(e) Given a, b, c, the three sides : then 
tan Ja == V[sin (s — a) sin (s — b) sin (s — <?)/sin s]/sin (s — a), 
tan ^b = V [sin (s—a) sin (s — b) sin (s — c)/sin s]/sin (s — b), 
tan^-c= V[sin(5 — a) sin (s — b) sin (s — c)/sins]/sin (s — c). 
Check : one of Delambre's formulae. 

Since each of the half angles is positive and acute, the radical 
must be taken positive and there is no ambiguity ; but no tri- 
angle is possible unless 5, s — a, s-b, s — c be all positive. 



160 SPACE TRIGONOMETRY. [V, PR. 

(/) Given A, B, c, the three angles : 
then tan \a— V [sin E/sin (a — e) sin (b — e) sin (c — e)] 

/sin (a — e), 
tan \b — \l [sin E/sin (a — e) sin (b — e) sin (c — e)] 

/sin (b — e), 
tan \c — V [sin E/sin (a — e) sin (b — e) sin (c — e)] 

/sin (c — e). 
Check : one of Delambre's formulas. 

Since each of the half sides is positive and acute, the radicals 
must be taken positive and there is no ambiguity ; but no tri- 
angle is possible unless e, a — e, b — e, c — e be all positive. 

questions. 

Solve these oblique triangles, given : 

1. a, 100°; b, 50°; c, 60°. 

[138° 15' 45", 31° 11' 14", 35° 49' 58". 

2. A, 120°; B, 130°; c, 80°. 

[144° 10' 2", 148° 48' 46", 41° 44' 15". 

3. I, 98° 12'; c, 80° 35'; A, 10 c 16'. 

[149° 32' 51", 30° 20' 29", 20° 22' 7". 

4. A, 135° 15'; c, 50° 30'; b, 69° 34'. 

[50° 6' 16", 120° 41' 47", 70° 28' 9". 

5. a, 40° 16'; b, 47° 14'; A, 52° 30'. 

6. a, 120°; b, 70°; a, 130°. 

[58° 57' 20", 75° 36' 4", 56° 13' 23", 
or 165° 23' 44", 163° 26' 16", 123° 46' 37". 

7. a, 40°; b, 50°; A, 50°. 

[65° 54' 52", 82° 48' 42", 56° 81' 24", 
or 114° 5' 8", 22° 16' 52", 18° 33' 2". 

8. A, 132° 16'; b, 139° 44'; «, 127° 30'. 

[136° 8' 16", 114° 17' 48", 77° 43' 4". 

9. A, 110°; B, 60°; a, 50°. 
10. a, 70°; B, 120°; a, 80°. 

[114° 49' 26", 65° 48' 58", 72° 56' 48". 



3,4,§15.J SOLUTION OF IDEAL OBLIQUE TRIANGLES. 1G1 

PROB. 4. TO FIND THE AREA OF AN IDEAL TRIANGLE. 

It is shown in geometry that the area of an ideal spherical 
triangle bears the same ratio to the area of a trirectangular 
triangle as the spherical excess bears to a right angle : 
i.e. if abc be a limited spherical triangle, and if k stand for 
the area of the triangle, t for the area of the trirectangular 
triangle, and 2e for the spherical excess a+b+c— 2r, then 
k:t=2e:b and k = t-2e/r. 
This area may also be expressed in terms of a, b, c, the sides 
of the triangle, as follows : 
Divide the equation [theor. 12, note 2, form. 4. 

tan %s • tan ^(s — a) — cot \a • cot \ ( <7 — a) 
by the equation [theor. 12, note 2, form. 3. 

cot ^(s — b) cot tt(s — c) — tan i(T cot \{a — a), 
then cot 2 \a — tan \s tan \{s — a) tan J(s — b) tan ^(s — c) ; 
and v 2e = ( a + b + c) - 2r = 4r - (a + ft + y) = 4r - 2&, 
.*. Je = r — %g, and tan^E = cot J<7, 
.*. tan 2 Je = tan \s tan ^(s — a) tan J(s — b) tan|-(s — c), 
.*. K=:T-4[tan _ V^ n ^tanJ-(5 — a) tanj(s — b) tan-J(s — c)] 

/K. 
Manifestly the radical has the positive sign for an ideal tri- 
angle, and the angle is the smallest positive angle in the group 
of congruent angles shown by the bracket. 



162 SPACE TRIGONOMETRY. [V, 

§ 16. RELATIONS OF PLANE AND SPHERICAL TRIANGLES. 

After the definition of the trigonometric ratios and the state- 
ment of their relations, all the properties of the right spheri- 
cal triangle, and of the plane triangle (oblique and right), may 
be derived from those of the oblique spherical triangle. Such 
a development of the subject presents the principles of trigo- 
nometry in their most general form, and teaches the student 
to take these general propositions and, by successive steps, to 
draw out and state in their logical order the special proposi- 
tions that are included in them. This mutual relation of the 
general and the particular not only helps the intellect to grasp 
these propositions, but also helps the memory to retain them. 

The order of development is this : 
to state and prove the general properties of the oblique spher- 
ical triangle, counting it the most general form of the 
triangle, 
to derive the properties of the right spherical triangle, count- 
ing it a special case of the oblique spherical triangle 
wherein one angle is a right angle, 
to derive the general properties of the oblique plane triangle, 
counting it a special case of the spherical triangle 
wherein the radius of the sphere has become infinite, 
to derive the properties of the right plane triangle, counting 
it a special case of the oblique plane triangle wherein 
one of the angles is a right angle, or of the right spher- 
ical triangle wherein the arcs are straight lines. 

The general properties of the plane triangle may be got from 
those of the spherical triangles as follows : 

If the sides of a spherical triangle subtend very small angles 
at the centre of the sphere, the spherical triangle differs but 
little from a plane triangle having the same vertices ; and, if 
the vertices be fixed in position while the centre of the sphere 
recedes further and further away, and the radii grow longer 
and longer, then the bounding arcs grow straighter, the spher- 
ical triangle approaches closer to the plane triangle having the 



§16.] RELATIONS OF PLANE AND SPHERICAL TRIANGLES. 163 

same vertices, the small angles at the centre of the sphere sub- 
tended by the sides of the triangle are proportional to those 
sides, and the sum of the three angles of the triangle is a little 
greater than, but approaches, two right angles. 

The plane triangle that has the same vertices is the limit to 
which the spherical triangle approaches when the radius is in- 
finite, and if in the formulae for spherical triangles the func- 
tions of the sides be expressed in terms of their subtended 
angles, and only those terms be retained whose limiting ratios 
are finite, i.e. those that are of the same low T est order of infini- 
tesimal, the resulting formulae correspond to the formulae for 
plane triangles. 

In detail : replace sin a, cos a, tan a • • • by 

a-a*/3!+ ■ ■ ., l-a 2 /2!+ ■ • -, a + a*/3 + ■ • - ; 
omit all terms except those of lowest order, and replace those 
infinitesimals by the corresponding sides of the plane triangle. 

(a) TJie terms of lowest degree of the first order : 
Eeplace sin a, tan a, 2 sin \a • • • by a ; cos a by 1 ; and so on ; 
then*/ sin a — sin c sin A = tan b cot B, 

:.a — c sin A = 5 cotB ; 
and v cos A = cos a sin B = tan b cot c y 

.*. cos a = 1 • sin B = b/c ; 
and v sin a/sin b = sin A/sin b, 

.-. a/b = sin A/sin b ; 
and v sin|A= V[sin (s — b) sin (s — c)/sin b sine], 

.*. sin -J a = V[(s — V) {s — c)/bc\ ; 
and v cos-^A= V[sins sin (s — a)/smb sine], 

.*. cos |a = V [s(s - a) /be] ; 
and v tan£A= \/[sin (s — b) sin (s — e)/sins sin (s — a)], 

.'. tan Ja = V [{s - V) (s - c)/s (s - a)] ; 
and v sin £(a - b)/cos |c = sin %(a — #)/sin \c, 

.'. sin \{ a - b)/cos \ c = (a — b)/c ; 



164 SPACE TRIGONOMETRY. [V, TH. 

and v cos £(a — B)/sin Jc = sin %(a + #)/sin \c, 

.'. cos |(a — B)/sin f c = (a + #)/e ; 
and v tan |(a — B)/cot Jc = sin £(a — J)/sin -£(« + 5), 

. \ tan £( a — B)/cot Jo = (a— i)/(a + 6) ; 
and y tan \{a + &)/tan j£= cos J(a — b)/cos J( a + b), 

.*. (a + b)/c = cos J(a — b)/cos^(a + b) ; 
and v tan %{a — S)/tan \c == sin |(a — B)/sin £(a + b), 

.\ (a — J)/c = sinJ(A — B)/sin J(a + b). 

(b) The terms of lowest degree of the second order : 
Keplace sin a, tan a, by a ; cos a by 1 — ^a 2 ; and so on. 
E.g. the formula cos c = cos a cos 5 becomes 

1_^ + ...=(1_^ + ...)(1_^ 2+ ...) 

.". a* + b* = c* ± terms of higher degree, whose ratios to 

a?, If, c 2 = when a, b, c = 0, 
.-.a 2 + b 2 =c 2 . 
So, the formula cos a = cos i cos £ +_sin b sin c cos a gives 
- a? = b 2 + c 2 -2bccosA. 

QUESTIONS. 

1. Show that, for a plane right triangle, of exs. 1-7, § 13, 

exs. 1, 2 reduce to a 2 -f W ■— & ; 

exs. 3, 4, to tan |a = V [(c - &)/(c + b)] = a/(c + 5) ; 

ex. 5, to a — b — a tan J a — b tan Jb ; 

exs. 6, 7, to a + b = 90°. 

2. Show that, for a plane triangle, of exs. 1-12, § 7, 

ex. 1 reduces to ex. 4, III § 2 ; 

ex. 3, to ex. 7, III § 2 ; 

ex. 5, to ex. 5, III § 2 ; 

ex. 7, to (5 - c)/(s — a) = tan J A/tan JC; 

3. Show what the other examples of § 7 reduce to. 



22, §17.] legexdre's theorem. 1G5 

§17. LEGEXDRE'S THEOREM. 

Theor. 22. If abc be any spherical triangle whose sides are 
very small as to the radius of the sphere, and if a'b'c' be a plane 
triangle tvhose sides a, b', c' are equal in absolute length to the 
sides a, b, c of the spherical triangle ; then each angle a, B, c 
exceeds the corresponding angle a', b', c' by one-third of the 
spherical excess of the triangle abc. 

For, replace sinJ, sine by b — ^b z + • • •, c — ^c z + • • •, and 

cosa, cosi, cose by l—%a 2 + • ■ •, 1 — J5 2 + • ■ •, 1— %c 2 + • • • ; 

thenv the formula cos a — cos b cos c + sin b sin c cos A gives 

be (cos a' — cos a) [cos a' = (b 2 + & — a 2 )/2bc. 

= ^(a 2 b 2 + b 2 c 2 + c 2 a 2 ) - j\ (a' + V + c 4 ) ± terms whose 

ratios to these terms = 0, when a, b, c = 0, 

and so for ca (cos b' — cos b), ab (cos c' — cos c), [sym. 

.*. be (cos a' — cos A) = ca (cos b' — cos B) = ab cos c' — cos c). 

But vcos a' — cos A = 2sin|-(A — a') sin^A + A') 

= (a — a') sin a', 
and so for cosb' — cos b, cose' — cose; [sym. 

.'. be (a - a') sin a' = ca (b - b') sin b' = ab (c - c') sin c ; 
and v sin a', sin b', sin c' are proportional to a, b, c, 

. .'. A-a'=B-B' = C-C'^l[(A + B + C)-(A'-f-B / + C')]. 
QUESTIONS. 

1. Triangles upon the earth's surface are regarded as spheri- 
cal triangles, and the earth's mean radius is 3956 miles. If 
the angles a, b be 65°, 60° and the side c be 100 miles, find the 
sides a, b in degrees and in miles ; find the angle c and the 
spherical excess ; find the area of the triangle in square miles ; 
find the number of square miles that correspond to 1" of 
spherical excess. 

2. In a geodetic survey the angles A, b, c are 48° 45', 30°, 
101° 15' 12", and the side c is 70 miles : find the angles of the 
plane triangle whose sides equal a, b, c, of the spherical tri- 
angle, and thence find the lengths of a, b. [Leg. th. 



1G6 SPACE TRIGONOMETRY. [V, 

§ 18. THE GENERAL SPHERICAL TRIANGLE. 

On page 116 it was shown that three lines, a'a, b'b, c'c, 
which meet in a point o, give four pairs of symmetric triedral 
angles, in the geometric sense : 

o-abc, o-a'b'c'; o-a'bc, o-ab'c'; 

o-ab'c, o-a'bc'; o-abc', o-a'b'c ; 




and that when the planes of these lines are properly directed, 
each triedral gives eight spherical triangles, thus forming eight 
groups of eight, each of the sixty-four differing in some way 
from every other one of them. 

So, it was shown on page 124 that the law of cosines and the 
law of sines hold true for all spherical triangles and all trie- 
drals, without regard to the signs or magnitudes of their parts ; 
and consequently that all the formulae derived from these laws 
hold true universally. 

In these triangles the circuit is so made that the vertices 
are always taken in the order a-b-c-a-b, never in the order 
a-c-b-a-c : that order would give sixty-four more triangles. 

It remains to show how, starting with one of them, e.g. the 
ideal triangle, the other sixty-three triangles may be derived 
from it ; and how, in the solution of the general spherical tri- 
angle, the species of the parts may be known. 



;18.] THE GENERAL SPHERICAL TRIANGLE. 



1C7 











108 SPACE TRIGONOMETRY. [V, 

THE DEFORMATION OF SPHERICAL TRIANGLES. 

Let abc be an ideal spherical triangle, as shown in the first 
figure, with the parts a, b, c, a, fi, y named and used 
as on page 112 ; 

let the plane a turn upon one of its own diameter, e.g. on b'b 
or on c'c, as upon a hinge, till it comes again to pass 
through the points b, c, but has its direction reversed, as 
in the second figure ; 

then, while the arcs b, c are unchanged, the arc a is replaced 
by its explement, 2n — a ; 

and v op, the axis of the plane a, is reversed with the plane, 
.*. the angle op makes with oq, the axis of the plane b, is 
made greater or less than before by two right angles, 
.\ the diedral a'b = y±7T, [a'= the reversed plane a. 

the diedral ca ' = (3±7t, and the parts of the new triangle are 

%n — a, b y c, a, fi±7t, ydz7T. 
So, if the plane b be reversed, and the planes c, a not, the parts 

of the triangle so found (third figure) are 

a, 27r — b, c, a±7t, /?, y±.n. 
So, if the plane c be reversed, and the planes a, b not, the parts 

of the triangle so found (fourth figure) are 

a, b, 27T—c, a±7t, /3±7t, y. 
So, if the two planes b, c be reversed, and the plane a not, the 

parts of the triangle so found (fifth figure) are 

a, 27t — b, %7t — c, a, /3±7t, y±n. 
So, if the two planes c, a be reversed, and the plane b not, the 

parts of the triangle so found (sixth figure) are 

2n — a, b, %7t — c% a±7t, ft, y±7t. 
So, if the two planes a, b be reversed, and the plane c not, the 

parts of the triangle so formed (seventh figure) are 

2n — a, 27t — b, c, a±7T, fi±7t, y. 
So, if the three planes a, b, c be all reversed, the parts of the 

triangle so found (eighth figure) are 

2n — a, %7T — b, 27C — C, a, fi, y t 



§ 18.J THE GENERAL SPHERICAL TRIANGLE. 169 

Let the line a have its direction reversed and a' become the 

positive end ; 
then the triedral o-a'bc gives eight triangles, which may be 

got from those of the triedral o-abc by noting, that : 
with the three planes a, b, c fixed, the diedrals ca, ab, and. the 

face angle boc are unchanged, 
the diedral be is reversed, being viewed from the other end of a, 

the co-line of the two planes b, c, 
and the face angle coa is replaced by coa ; , and aob by a'ob. 

E.g. the parts of the triangle that mates with the ideal triangle 

are a, b±.n, c±.n, —a, ft, y ; 
and the other seven may be got in like manner. 

So, if the line ft be reversed, or the line y, there are eight new 
triangles symmetric with the eight triangles of o-a'bc. 

If the two lines ft, y be reversed, and a not ; 

then the diedral be is unchanged, the diedrals ca, ab are re- 
versed, the face angles coa, aob are replaced by c'oa', 
aob', and b'oc' is equal to boc. 

E.g. the parts of the triangle that mates with the ideal triangle 

are a, b±n, c±7t, a, —ft, —y ; 
and the other seven may be got in like manner. 
So, -if the two lines y, a be reversed, or the two lines a, /3, 

there are eight new triangles symmetric with the eight 

triangles of o-ab'c'. 

If the three lines a, ft, y be all reversed ; 

then, for any one position of the three planes the diedrals are 
all replaced by their opposites, and the new face angles 
b'oc', c'oa', a'ob' are equal to boc, coa, aob. 

E.g. the parts of the triangle that mates with the ideal triangle 
are a, b, c, — #, — ft, —y. 

The parts of thirty-two of these sixty-four triangles are 
shown in the table below, and the rest may be written by sym- 
metry. In this table the angles are all set down as positive, 
the negative angles — a, a — n* • • being replaced by their 
next greater positive congruents. 



1T0 SPACE TRIGONOMETRY. [V, 

These triangles have all their parts positive and less than 
four right angles, and they are called the primary triangles 
of the three co-pointar lines a, fi y y, always naming the lines 
in this order. A triangle that has parts negative or greater 
than four right angles may be reduced to one of these by add- 
ing or subtracting multiples of four right angles. 



c 






2it—a, b, c, 


a, 2it—b, 2it—c, 








a, 7t + fi, 7t + y. 


a, 7t + ft, 7t + y. 


a, 


h 


c, 


a, 2it—b, 7t + y, 


2it—a, b, 2it — c, 


tf, 


ft, 


r- 


n+a, ft, n+y. 


7t+a, ft, 7t + y. 



27t—a, 27t—b, 2it—c, a, b, 2n—c, 2n—a, 2n—b, c, 

a, ft, y. 7C + a, n + ft, y. it + a, 7t + fi, y. 

a'b'c' 2it—a i b, c y a y 2it—b,2it — c, 

2n—a, it— ft, rt—y. 2it—a, it— (5, n—y, 

a, b, c, a, 2it—b, c, 2it—a, b, 2it—c, 

2it—a,2it—ft,27t—y. Tt—a,2it—fi, n—y. Tt—a,2it—(5, it—y. 

27t—a, 2rc—b, 2it—c, a, b, 2it—c, 2it—a, 2n—b, c, 

27t—a,27t—ft,27t—y. it— a, it—fi,2it—y. it— a, it— ft,27t— y* 

a' b c 2it— a, it+b, rt + c, a, it—b, n—c, 

2it—a, 7t + ft, Tt + y. 27t—a, 7t + /3, 7t+y. 

7t + a, 2it—b, 7t + c, it— a, b, 7t + c, 

rt + a,27t—(3, itA-y. 7t + a,27t—fi, it\y. 

7t + a, 7t + b, 2-rt—c, it— a, 7t—b, c, 

7t+a, it + ($,2it—y. n+a, 7t + ft,27t—y. 

ab'c' %7t— a, 7t + b, 7t + c, a, 7t—b, n—c, 

a, 7t — fj, it—y. a, it— ft, it—y- 

a, 7t + b, 7t + c, 7t + a, 2it—b, 7t-\-c, 

a, 2it—(5, 2it—y. it— a, /3, it—y. 

2it—a, it—b, rt—c, 7t + a, 7t + b,27t—c, 

a, 2n—ft, 2it—y. it— a, it— ft, y. 



a, 


it + b, 


7t + C, 


2n—a, 


ft, 


r 


27t—a, 


7t—b, 


Tt — C, 


2it—oc, 


ft, 


r 



it— a, 


b, 7t—C, 


it— a, 


ft> n-Y' 


it— a, 


n—b, c, 


it— a, 


*-ft, r 



§18.] THE GENERAL SPHERICAL TRIANGLE. 171 

DETERMINATION OF THE SPECIES OF THE PARTS. 

These sixty-four triangles have been divided into two classes 
called 'proper triangles and improper triangles. 

To the first class belong the ideal triangle ... 1 
and those got from the ideal triangle by reversing : 

one side or one angle, 6 

one side and the opposite angle, 3 

two sides and one opposite angle, 6 

two angles and one opposite side, 6 

two sides and the two opposite angles, ... 3 

the three sides and two angles, 3 

the three angles and two sides, 3 

all the sides and angles, 1 

in all, thirty-two proper triangles. 32 

The other thirty-two triangles are improper triangles ; and 
it will appear that the upper signs' of Delambre's formulae must 
be used in solving proper triangles, and the lower signs in solv- 
ing improper triangles. 

Certain limitations must also be observed, as below : 
Let a, b, c, a, /3, y be the parts of the ideal triangle, 
and a r , I)', c', a! , /?', y' the parts of any primary triangle ; 
let s'^a' + b' + c') and <?' = ±(a' + fi' + y') ; then: 

1. If the data make a solution possible, the products 

sin s' sin (s r — a') sin (s r — V) sin (s' — c r ), 
sin 6 sin((T' — a') sin (a' — /?') sin (a' — j/), 
are both positive, since they are perfect squares, [th. 5, cr. 3. 

2. s'= (s'-a f ) + (s'^b'y *(a^), 

3. The sixty-four triangles show but ten distinct type-forms, 

so far as concerns their sides : 

a' —a, a 1 — a, a' — a, a' — a, a' = a, 

b' = b, b f = 27t-b, b' = 7t + b, b'=7t + b, V-n-b r 

c' =c; c' — %7t — c] c'=7t + c; d—n — c\ c'—n — c; 



172 



SPACE TRIGONOMETRY. 



[V, 



a—%7t — a, a' = 27t — a 3 a! — 2n — a, a 1 — 2n — a i d = 27t~a, 

b' = b, l)' = 27T-b, b' = 7r + b, b' = 7t + b, b f =7t-b, 

c —c\ d — 2n — c\ d—7t-\-c\ c r =7t — c; c r =7r — c; 

and there are eight possible groups of consistent inequalities: 



o <y< n, o <s 

27T<s'<37r, <s 

7r<s'<27T, n<s 

7T<S' <27t,—7t<S 

7T<S'<27T, <S 

7r<s'<27r, <s 

7T<s' <2n, <s 



-a'<7T, <s'-b'<7t, <s'-c'<7t ; 

-a'<7t, <s f -b'<7r, <s' -c'<7t. 

a'<27T, 0<s'-b'<7r, <s'-c r <7t ; 

■a'<0, <s'-b'<7t, 0<s'-c f <7r. 

-a'<.7t, 7T<s' — b' <2zr, <s' — c <7i ; 

-a'<7r,-7r<s r -I' <0 , <s' — c'<7t. 

-d<7t, <s' -V <n, 7t<s' — c f <27r; 
7t<s'<27t, <s'-a'<7t, <s' -b'<7r,-7t<s f -c'< 0. 

with like type-forms and like groups of the angles. 

These inequalities are relied on to determine the species of 
the parts sought. There are always two solutions : real and 
separate, real and coincident, or imaginary. 

fi.g. in case (a), given b,c, a : the angles i(fi + y), \(fi — y) 

are both two valued. 
But0<^-(/?-hx) <2tt, -7t<±(/3-y) <7t; and together they 

give only two values each to fi, y between and 2n ; 
and a is found from the values of /?, y without ambiguity. 

So, in (c) y has two values, and a, a have single values for 
each value of y. 

So, (b, d) follow (a, c), and (e, f) show no ambiguity. 

GRAPHICAL SOLUTION. 

The graphical solution of a primary triangle is made in the 
same way as that of the ideal triangle, angles greater than two 
right angles being replaced by their next less negative con- 
gruents. If any construction be possible, there are two con- 
structions, which may be separate or coincident. 



§ 19.J SPHERICAL ASTRONOMY. 173 

§ 19. SPHERICAL ASTRONOMY. 
THE CELESTIAL SPHERE. 

In astronomy the elements of position of a heavenly body are 
distance and direction ; in spherical astronomy only one of these 
elements, direction, is regarded, and that is usually referred to 
the earth's centre. For this purpose all stars may be considered 
as at the same distance from the earth's centre upon the sur- 
face of a sphere of arbitrary radius called the celestial sphere. 

The trace of the plane of the earth's equator on this sphere 
is the celestial equator, whose poles (north and south) are the 
traces of the earth's axis. 

The ecliptic is a great circle of the celestial sphere, the sun's 
apparent path in one year due to the motion of the earth around 
the sun ; it cuts the equator in two points, the vernal and the 
autumnal equinox, which are passed through by the sun, about 
March 20 and September 23. The obliquity of the ecliptic is 
the nearly constant angle of 23° 27' between the planes of the 
ecliptic and equator. 

Secondaries to any great circle, or primary, are great circles 
cutting it and therefore its parallels at right angles. Second- 
aries to the celestial equator are hour-circles or meridians. 

To any observer the sensible horizon is a plane touching the 
earth's surface at the point of observation ; and a plane parallel 
to this plane through the earth's centre traces out on the celes- 
tial sphere the rational horizon, whose poles, zenith and nadir, 
are the traces of a vertical line, and whose secondaries are ver- 
tical circles. One of the vertical circles is also an hour-circle, 
the observer's celestial meridian, and passes through his zenith 
and nadir, and the north and south poles of the celestial sphere ; 
its plane is the same with that of the observer's terrestrial me- 
ridian, and it meets the plane of his sensible horizon in his 
meridian line. The vertical circle that is perpendicular to the 
meridian is the observer's prime vertical, and it goes through 
the east and west points of his horizon. 



174 SPACE TRIGONOMETRY. [V, 

SPHERICAL COORDINATES. 

As the position of a point on the earth's surface is defined by 
means of two coordinates (latitude and longitude), the stan- 
dards of reference being a convenient great circle (the equator) 
and a convenient point on it (the point where it is crossed by 
the meridian of Greenwich) ; so, the position of a star at any 
instant on the celestial sphere may be defined in either of three 
ways : 

1. As to the celestial equator : 

The decimation of a star is its angular distance (north or 
south) from the celestial equator measured upon its hour-circle ; 
and the arc of the equator intercepted between this circle and 
the vernal equinox is the star's right ascension ; it is reckoned 
eastward from the vernal equinox from 0° to 360°. The com- 
plement of the declination is the polar distance. 

Instead of a star's right ascension its hour-angle is often used, 
in problems that involve diurnal motion, to define its hour- 
circle at any instant ; this angle is the angle at the pole between 
the observer's celestial meridian and the star's hour-circle, and 
is counted from the meridian, positive tow T ards the west and 
negative towards the east. The right ascension of a fixed star 
changes very little, since the vernal equinox is nearly fixed on 
the celestial sphere ; the hour-angle changes every moment. 

2. As to the ecliptic : 

The latitude of a star is its angular distance from the ecliptic 
measured on a secondary ; and the arc of the ecliptic inter- 
cepted between the vernal equinox and this secondary, meas- 
ured eastward, is the star's longitude. 

3. As to the horizon : 

The altitude of a star is its angular distance from the horizon 
measured on a vertical circle ; and the arc of the horizon inter- 
cepted between this circle and the south point of the horizon 
is the star's azimuth. Owing to the rotation of the celestial 
sphere, the horizon-coordinates change every moment. 



§ 19.] SPHERICAL ASTRONOMY. 175 

RELATIONS BETWEEN ECLIPTIC-COORDINATES AND EQUATOR- 
COORDINATES. 

On the celestial sphere let p be the pole of the equator, Q the 

pole of the ecliptic ; 

then the great circle through p, q is the common secondary of 

the equator and the ecliptic. 
Let v, w be the vernal and the autumnal equinox at quadrantal 

distances from s, t ; 
let pam be the hour-circle of a star A, and qan the secondary 

to the ecliptic ; 
then vm, ma are the right ascension and declination of A, and 

yn, na are its longitude and latitude. 



These four coordinates of any fixed star are subject to only 
slight variations in any one year ; they are recorded for the 
principal stars in a yearly almanac, with the data for computing 



176 SPACE TRIGONOMETRY. [V, 

the variations ; the sun's declination is recorded for each day or 
half day, and may be got for any hour and minute by interpo- 
lation. 

The spherical angle xvs is the obliquity of the ecliptic, and, 
since yx, ys are quadrants, xys is measured by the arc xs; 
arcs xs, PQ, yt are each 23° 27', and arcs sp, yq are each 66° 33'. 

Equator-coordinates may be converted into ecliptic-coordi- 
nates : 
when ym, ma are given in the right spherical triangle mya, 

the arc ya and the angle mya may be found ; 
the angle kya is found by subtracting the obliquity, and the 

triangle nya may be solved for yn, na ; so conversely. 

the sun's annual motion. 

The particular case of the sun is simpler : since his apparent 
annual path is the ecliptic, his latitude is always zero, and his 
right ascension, declination, and longitude are the arc-abscissa, 
arc-ordinate, and arc-distance of a given angle, the obliquity ; 
his declination increases from 0° at y on March 21 to 23° 27' 
at s on June 21 (the summer solstice), then decreases to 0° at 
w on September 21, and to "23° 27' at t on December 22 (the 
winter solstice), then increases to 0° at y ; his right ascension 
and longitude are equal at 0°, 90°, 180°, 270°, 360°. 

QUESTIONS. 

1. The altitude of a circumpolar star at upper transit across 
meridian is 60°, and at lower transit 40° : find the declination 
of the star. 

2. The vernal equinox culminated (reached its highest point) 
at h 10 m 13 s , and a certain star culminated at 2 h 5 m 10 s : find its 
right ascension. 

3. Find the latitude and longitude of a star whose right 
ascension is 5 b 13 m , and declination 60°. 

4. When the sun's declination is 15°, find his right ascen- 
sion and longitude. 



5 19- J 



SPHERICAL ASTRONOMY. 



177 



RELATIONS BETWEEN EQUATOR-COORDINATES AND HORIZON- 
COORDINATES. — THE ASTRONOMICAL TRIANGLE. 

On the celestial sphere let p be the pole of the equator xy, 
and z that of the horizon ns ; 

then the great circle through pz is the celestial meridian, the 
common secondary of equator and horizon ; 

let zwz'e be the prime vertical perpendicular to both meridian 
and horizon and meeting both equator and horizon in 
the east and west points. 




The celestial sphere appears to make a complete revolution 
on its axis pp' in about 23 h 56 m 4 s of civil time. This is the 
interval between two successive transits of any fixed star, and 
is a sidereal day. A sidereal clock shows hours when the 
vernal equinox culminates ; and the hours are marked from 
to 24. The sidereal time of a star's transit gives its exact 



178 SPACE TRIGONOMETRY. [V, 

right ascension, which may be converted into angular measure 
at the rate of 15° to a sidereal hour, or 1° to 4 minutes, 15' to 
1 minute, 1' to 4 seconds, and so on. 

The hour-circle of the star A coincides with the meridian in 
the position pa bearing due south as seen from o, and the star 
has then its greatest altitude : 

in the position pa 2 the star is on the prime vertical and bears 
due west ; 

in the position pa 3 the star sets below the horizon ; 

it reaches its greatest depression at a 4 when its hour-circle 

passes over the meridian bearing due north ; 
it rises at a 5 , reaches the prime vertical at A 6 bearing due east, 

and culminates again at A . 

The spherical triangle zpa x for any position of the star A is 
the astronomical triangle : 

its sides za 1? pa x are the co-altitude and co-declination of A x ; 
the angles zpa 1? pza x are the supplement of the hour-angle and 

of the azimuth of A ; 
and the side pz is the observer's co-latitude. 

For v this co-latitude is the angle between the earth's axis and 

the vertical line at the point of observation, 
and the traces of these lines on the celestial sphere are p, z, 
.*. the arc pz measures the observer's co-latitude. 

When the latitude is known the relations between the sides 
and angles of this triangle give the relations between the star's 
equator- and horizon- coordinates. 

The observers latitude may be determined, once for all, by 
the astronomical triangle when the declination, the altitude, 
and either the azimuth or the hour-angle of a heavenly body 
are known for some instant. If at the time of observation the 
body be on the meridian, the hour-angle is zero, and the azi- 
muth either zero or 180° ; if it be on the prime vertical, the 
azimuth is ± 90° ; if it be on the horizon, the altitude is zero ; 
and in all these cases the computation of latitude is simplified. 



§10.] 



SPHERICAL ASTRONOMY. 



179 



THE SUN'S DIURNAL MOTION. — SOLAR TIME. 

The sun's hour-circle ps coincides with the observer's merid- 
ian at noon ; the hour-angle zps x at any instant measures the 
time of observation from noon ; the angle zps 3 measures the 
time of sunset. 

The greater the declination xs , the greater is the hour- 
angle of setting, zps 8 , the longer the day, and the shorter the 
night. The day is longest in the northern hemisphere when 
the declination is greatest, June 21, the summer solstice. 

z 




When the declination is zero, the diurnal path s s 5 coincides 
with the equator and is bisected by the horizon ; the day is 
then equal in duration to the night, and hence the term equi- 
nox. When the declination is 23° 27' s., the day is shortest 
in north latitudes, and the night longest (winter solstice). 

The interval between two successive transits of the sun over 



180 SPACE TRIGONOMETRY. [V, 

the same meridian is an apparent solar day. This interval 
varies, from two causes : the obliquity of the ecliptic, and the 
variability of the sun's apparent motion in the ecliptic. 

The mean sun is an imaginary body, supposed to move uni- 
formly in the equator with the annual period, and with the 
average velocity, of the true sun. It culminates at civil or mean 
noon, and the constant interval between two successive transits 
is a mean solar day. This interval is divided into hours, min- 
utes, and seconds. A second of mean solar time is the ordinary 
time-unit, and is the same fraction of a mean solar day as a 
sidereal second is of a sidereal day. The mean solar time is the 
hour-angle of the mean sun, at any instant ; the apparent solar 
time is the hour-angle of the true sun. The angle between 
the mean and true hour-circles is recorded for each day, in 
the almanac, as the equation of time. It varies throughout the 
year between and about ± 16 minutes of time. 

The astronomical day begins at mean noon, and the hours 
are numbered from to 24. 

In what follows, apparent time is used. 

QUESTIONS. 

1. The meridian altitude of the sun's centre was 25° 38' 30" 
S., and his declination 22° 18'' 14" s. : find the latitude. 

2. The meridian altitude of Jupiter was 50° 20' 8" s., and 
his declination 18° 47' 37" N". : find the observer's latitude. 

3. The sun crossed the prime vertical at an altitude of 54°: 
find the observer's latitude and the time of day, the sun's dec- 
lination, got by interpolation for the approximate time of day, 
being 18° 30'. 

Here, zs 2 = 36°, ps 2 = 71°30', pzs,=90° : find zp, zps 2 . 

4. Find the observer's latitude in ex. 1, page 176. 

In what latitude will this star just graze the horizon ? 

5. If the sun's declination be 22° 26' ^., and altitude 40° 55' 
at 3 p.m., find the observer's latitude. 

In this example, zs 1 = 49°5', ps 1 =t67°34', ZPS! = 3 h. = 45°, 
and the co-latitude, pz, is to be found. 



§19.] SPHERICAL ASTRONOMY. 181 

6. In latitude 13° 17' K. the sun's altitude was 36° 37', his 
declination was 22° 10' s. : find his hour-angle. 

7. If the sun's declination be 17° isr., find the time in the 
afternoon when he will be due west from a place in latitude 51° 
N.; and find how far from the west point he will set (his am- 
plitude at setting). 

8. If the sun be due west at setting (amplitude zero), find 
his declination and the time of year. 

9. If the time of sunset be sought on any given day, a quad- 
rantal triangle pzs 3 may be solved for the hour-angle zps 3 . 

If the sun's declination be 14° s. and the latitude 42° N., find 
the time and amplitude of sunrise and sunset. 

10. Find the time of setting in ex. 8. 

11. Find the length of the longest day in Ithaca, excluding 
twilight, latitude 42° 30' N. 

12. Find the lowest north latitude in which the sun does 
not set on the longest day, nor rise on the shortest day. 

13. Find the time of sunrise in Boston, latitude 42° 21' n., 
on the shortest day of the year, and the sun's amplitude. 

14. The phenomenon of twilight is due to the reflection and 
refraction of some of the sun's rays toward the observer's eye 
when the direct rays are intercepted : it begins or ends when 
the sun is about 18° below the horizon. 

How long does twilight last in Boston on the shortest day ? 
Given zs 4 =90° + 18°, ps 4 , zp : find zps 4 , and subtract the 
hour-angle of sunset, zps 3 . 

15. Find the length of the longest day in Ithaca, including 
morning and evening twilight. 

16. In what latitude does the sun get just 18° below the hori- 
zon on the longest day, so that twilight lasts all night ? 

Here, ns 5 = 18°, ps 6 = 66° 33': find the co-latitude zp. 

17. Given the declination of Aldebaran, 1 6° 17' N. : find his 
altitude and azimuth to an observer at Boston when the hour- 
angle of this star is 3 h 25 m 12 s ; and find the hour-angle and am- 
plitude at rising and setting. 



182 SPACE TRIGONOMETRY. [V, 

§20. NAVIGATION. 

When a mariner cannot make celestial observations, he has 
recourse to dead-reckoning ; i.e. he computes the position of 
his ship from the latitude and longitude of her starting-point 
or of the place of last observation, and the records of sailing. 
This dead-reckoning is the subject of navigation proper, as 
distinguished from nautical astronomy. 

The rate of sailing is usually recorded every hour, and is 
measured by the log-line. This is a line wound on a reel and 
attached to a small quadrantal piece of board. The quadrant 
is loaded on the arc with lead to keep it upright when thrown 
into the water and prevent its moving forward toward the ship 
while the line is running out. The log-line is divided into 
knots, each a hundred- twentieth part of a nautical mile, so 
that the number of knots run out in half a minute gives the 
ship's hourly rate in miles. 

Bearings at sea are given in points and quarter-points, 
counted from each of the eight cardinal points, two points 
each way. 




The direction of sailing at any time is shown by the mariner's 
compass. 

The reading of the compass is to be corrected for variation, 
deviation, and leeway. 

The variation is the angle between the magnetic and true 
meridians ; it is found, for various places, by astronomical ob- 
servations, and laid down on the nautical charts. 



§20.] NAVIGATION. 183 

The deviation is the angle of deflection of the needle from 
the magnetic meridian, caused by the iron of the ship ; it is 
found, for a given ship and a given direction, by special ex- 
periments. 

When there is a side wind, the angle which the ship's track 
makes with her fore-and-aft line is the leeway : it is found, for 
a given ship, a given freight, and a given obliquity and velocity 
of the wind, by special experiments. 

The corrected reading is the coarse ; it is the angle between 
the ship's true meridian and her true direction of motion. In 
what follows, the corrections are supposed to have been made, 
so that the given courses are the true courses. When the 
course is kept constant, the ship's track crosses every meridian 
at the same angle ; the path is neither straight nor circular, 
but a spiral, the loxodrome or rhumb-line, that goes round and 
round the earth's surface, coming nearer and nearer to the 
pole ; and its length is the distance. 

The meridian length between the first and last parallel of 
latitude is the difference of latitude made by the ship. 

The departure is her easting or westing from her first me- 
ridian ; it is measured as follows : if she sail on a parallel of 
latitude, the departure is the distance made on the parallel ; 
if she sail on a loxodrome, the departure for each successive 
instant is measured on the parallel she is then crossing, and 
the limit of the sum of these infinitesimal departures is the 
total departure. 

The unit of length is the nautical mile, about 6080 feet, a 
sixtieth part of a degree of a great circle of the earth. Sixty 
nautical miles are a little more than sixty-nine statute miles. 

In what follows the earth is regarded as a perfect sphere. 
The error thus introduced is too small to be taken into account 
in any calculations whose data are derived from the log-line 
and compass. 



184 



SPACE TRIGONOMETRY. 



[V, TH. 



PLANE SAILING. — RELATIONS BETWEEN COURSE, DISTANCE, 
DIFFERENCE OF LATITUDE, AND DEPARTURE. 

Let ad be the rhumb-line, pa, pd the first and last meridians, 
pm, vn • • • meridians at equal small intervals ; 

p 

Departu: 




let m'm, n f n, • • • be the small arcs intercepted on successive 

parallels ; 
then the total departure from A to D is the limit of the sum 
m r m -f nn -}-■•••* when the meridians are taken very close 
together. [df. dep. 

The infinitesimal triangles Kmm\ mnn' may be treated as 
right plane triangles ; and since the course is constant they are 
similar. The elements of the motion are thus given by a series 
of infinitesimal right plane triangles, the sum of whose hypot- 
enuses is the distance, of whose bases is the departure, and 
of whose altitudes is the difference of latitude. These three 
sums, and the course, have the same relations to each other as 
the parts of any one of the elemental triangles ; whence they 
may be accurately represented by the parts of a right plane tri- 
angle. For this reason, although the sphericity of the earth 
is taken into account, the term plane sailing may be applied 
to any problem into which the difference of longitude does not 
enter ; and the solution is effected by the rules for the solution 
of right plane triangles. 



23, §20.] 



NAVIGATION. 



185 



PARALLEL SAILING.— RELATIONS BETWEEN A DISTANCE SAILED 
ON A GIVEN PARALLEL OF LATITUDE AND THE DIFFERENCE 
OF LONGITUDE. 

Theor. 23. The length of an arc of a parallel of latitude is 
the product of the length of the equatorial arc of the same num- 
ber of degrees by the cosine of the latitude of the parallel. 

For let p be a pole of the earth, c is centre, a, a' any two points 
on the equator, pa, pa' two meridians catting a parallel 
of latitude in L, l'; 

let o be the centre of the arc ll'; 




thenv arc ll': arc a a' = ol: ca [geom. 

= ol/cl = cos acl = the cosine of the latitude ; 
/. LL' = AA'-the cosine of the latitude. q.e.d. 

MIDDLE LATITUDE SAILING. — APPROXIMATE RELATION BE- 
TWEEN THE DIFFERENCE OF LONGITUDE AND THE DE- 
PARTURE ON A LOXODROME. 

The departure from A to D lies, in value, between AB and CD, 
and for short distances is nearly the same as the ship makes if 
she sail between the same two meridians on the mid-parallel ; 
i.e. -the parallel whose latitude is half the sum of the latitudes 
of A and D. Hence the departure from a to D is taken equal 



186 



/ 

SPACE TRIGONOMETRY. 



[V 



to the product of the difference of longitude of A and D by the 
cosine of their middle latitude. [theor. 23. 

The difference of longitude is thus connected with the other 
elements of the ship's path. 

MERCATOR'S PROJECTION. — ACCURATE RELATION BETWEEN 
THE DIFFERENCE OF LONGITUDE AND THE DEPARTURE 
ON A LOXODROME. 

Project the figure of page 184 on a plane surface as follows: 

C Difference of Longitude DC D 

G6~ 




28°45' 











/ 
















lV 




P 




nV 




n 




111/ 




111 







A60° 



1 . Draw a horizontal line for the equator, and vertical lines 
at equal intervals for the meridians. 

It follows that the projection m'm of any arc of a parallel is 
equal to the corresponding arc of the equator, and is therefore 
multiplied by a projecting factor, the secant of its own latitude. 

2. Draw a straight line cutting the meridians at the con- 
stant angle given by the course. 

It follows that the angles of each small plane triangle re- 
main the same ; so that while each triangle is enlarged, its 
shape is preserved, and m'n or mn' has the same projecting 
factor as mm \ and n'p r or np' the same projecting factor as 
n'n, and so on. 

Each small portion of the meridian in the neighborhood of 
any parallel is therefore multiplied by the secant of the lati- 
tude of that parallel, and the total length of the projection of 



§20.] NAVIGATION. 187 

any given portion of a meridian is the limit of the sum of 
these products, when the parts are taken indefinitely small. 
In practice it is sufficiently accurate to take each part as two 
minutes or nautical miles, and to use as its projecting factor 
the secant of the latitude of its middle point. 
E.g. the meridian-arc between the equator and latitude 13° 16' 
projects into a distance on the chart equal to the sum 

2 (sec 1' + sec 3' + sec 5' + • • • + sec 795') 
in nautical miles, on the assumed scale. 
This distance is computed and tabulated as the meridional 
part for 13° 16'. In computing such a table, each entry may 
be used in succession to find the next one, e.g. the meridional 
part for 36' is found from that for 34' by adding 2 sec 35'. 

The difference between the meridional parts for two lati- 
tudes is their meridional difference of latitude. 

In the figure above, AC is the meridional difference of lati- 
tude from a to D, and cd is the difference of longitude ; 
and dif. long./merid. dif. lat. =tan course 

= dep./true dif. lat. [plane sailing. 

These equations connect the difference of longitude with 
the other elements of the ship's motion : 
E.g. given the latitude and longitude of A, the course and dis- 
tance from a to D : 
find by plane sailing the departure, the difference of latitude, 

and the latitude of d ; 
find the meridional difference of latitude by subtracting me- 
ridional part for latitude a from that for latitude d ; 
compute the difference of longitude from the above relations. 
Note. The student of the calculus will see that the exact 
meridional part for latitude A is 

/* sec \-d\ — \og e tan (i7r + ^A), in radians ; 
and this result may be reduced to nautical miles, as follows : 

•.■log e tan(i^ + ^A) = log 10 tan(45 o -{-jA).2.3026, 
and r=3487.75' l 2.3026-3437.75 = 7916, 

.-. log, tan (45° + JA) = log 10 tan (45° + U) • 7916. 



18S SPACE TRIGONOMETRY. [V, PR. 

E.g. if A = 13° 15', 

then 45° + iA = 51° 37' 30", 

the merid. part = 0.10134- 7916 = 802 nautical miles, 

and this latitude is enlarged in the ratio 802 : 795. 

TRAVERSE SAILING. 

PROB. 5. To REDUCE THE RESULT OF SEVERAL SUCCESSIVE 
COURSES AND DISTANCES TO A SINGLE COURSE AND DISTANCE. 

(a) The latitude of the starting-point not given : 
Compute each separate difference of latitude and departure by 

plane sailing ; 
take the algebraic sum of the separate differences of latitude 

for the value of the direct difference of latitude, 
and the algebraic sum of the departures for an approximate 

value of the direct departure; 
find the direct course and distance by plane sailing. 

(b) The latitude of the starting-point given : 

Compute the separate differences of latitude by Mercator's or 
middle-latitude sailing ; 

take their algebraic sum for the direct difference of latitude, 
and so for the differences of longitude ; 

from these find the direct departure by Mercator's or middle- 
latitude sailing ; 

find the direct course and distance by plane sailing. 

Note. The first course and distance entered are usually got 
by taking a departure, i.e. by taking the bearing and distance 
of some object of known latitude and longitude ; the reverse 
of these are entered on the log-slate as the first course and dis- 
tance. 

GREAT CIRCLE SAILING. 

The shortest distance between two places is the great circle 
arc joining them ; it does not cut all the meridians at the same 
angle ; hence to keep on a great circle the ship must contin- 



5, §20.] 



NAVIGATION". 



189 



nally change her course. By means of a chart; several places 
on the great circle may be determined, and if the ship lay her 
course for these on successive rhumb-lines, her path will differ 
little from the circular arc. 

The elements of the great circle track between two given 
places are the distance, the first and last courses, and the high- 
est latitude passed through. These are got from the spherical 
triangle whose vertical angle at the pole is the difference of 
longitude of the two places, and whose sides are their co-lati- 
tudes. 

CURRENTS. 

In order to ascertain the set and drift of a current, i.e. its 
direction and velocity, a boat is taken a short distance from 
the ship and kept stationary by letting down a heavy weight; 
the log is thrown from the boat, and the direction in which it 
is carried, i.e. the set of the current, is taken by the boat com- 
pass, while the drift is given by the number of knots run off 
in half a minute. The effect of the current is considered 
equivalent to an independent course. 

E.g. if a ship sail 10 knots an hour in a current setting s.e. 5 
miles an hour, what course must she lay to make a place 
whose bearing is s. w. by s. ? 




(a) By construction and measurement. 
Take ab pointing s.e., and equal to 5 on any scale ; 
take AC pointing s. w. by s. ; 



190 SPACE TRIGONOMETRY. [V, 

with B as centre, and radius bc equal to 10, cut AC in the point 

c ; complete the parallelogram abcd : 
the angle sad is the course sought. 

(b) By computation. 
In the triangle abc, the sides ab, bc and the angle bac being 
known, compute the angle bca, and thence the course 

SAD. 

TACKING. 

A ship is on the starboard tack when the wind is on her 
right, on the port tack when the wind is on her left ; she is 
close-hauled on either tack when she sails as nearly as possible 
toward the point whence the wind blows. 

If when close-hauled she find her destination lying between 
her path and the wind, then she cannot reach it on this single 
tack ; but she may continue till the angle that the direction of 
her destination makes with the wind is just equal to her angle 
of close-haul, and then run in close-hauled on the other tack. 

E.g. if a ship can sail within 6 points of the wind on the port 
tack, and within 5J points on the starboard tack, 

find her course and distance on each tack to reach, in the short- 
est time, a point 15 miles N".w., with the wind due w T est : 

(a) By construction and measurement. 

Take AC pointing x.w., and equal to 15 on any scale; 

for the port tack draw ab 6 points to the right of the wind ; 

for the starboard tack draw ad 5J points to the left of the wind ; 

from the point c draw cb parallel to da ; 

measure ab and bc for the distances on each tack. - 

(b) By computation. 

In the triangle abc, ac = 15, A = 6 — 4= 2 points, 



B = 8-6-f-8-54- = 4i points, c = 5J + 4 = 9£ points, 
check : a + b + c = 16 points ; compute ab, bc 

The answer is the same whichever tack be taken first. 



•>o.] 



NAVIGATION. 



191 



Note. The surface of the earth is supposed to be flat within 
the limits of these problems. They come usually under case 
(b) in compound-course sailing. 




QUESTIONS. 

1. A ship sails due west 117 miles from a point in lat. 38° n. , 
long. 16° e. : find the longitude reached. [13° 31' 30" E. 

2. In what latitude is a degree of longitude half as long as 
at the equator ? 

3. Sails, e. 67 miles from New York light, lat. 40°28'n., 
long. 74° 8' w. : by middle-latitude sailing find the latitude and 
longitude of the point reached. [39° 40' 36" N., 73 d 6' 6" w. 

4. Find the course and distance from Montauk Point,41° 4'x., 
72° w., to Martha's Vineyard, 41 17'n., 70° 48' w. 

Her. 76° 30' 40" e., 55. 73 miles by middle-latitude sailing ; 
_jnv76°43'e., 56.58 miles by Mercators sailing. 

5. A ship sails from a point 14° 45' N., 17° 33' w., on a course 
s.28°7'30"w., till she reaches longitude 29°26'av. : find by 
Mercator the distance sailed and the latitude. 

[1500 miles, 7°18's. 

6. Prom a point in latitude 50° 10' s. a ship sails s. 67° 30' e. 
till her departure is 957 miles : find by Mercator the distance 
sailed, the difference of latitude, and the difference of longi- 
tude. [1036, 6° 36' 24", 26° 53'. 



192 SPACE TRIGONOMETRY. [V, § 20. 

7. A ship starting from a point in latitude 32° x. sails 
N. 25° e. 16 miles, thence s. 54° e. 11 miles, thence n*. 13° w. 7 
miles, thence ST. 61° E. 5 miles, thence K. 38° w. 18 miles : find 
the single course and distance that would bring her to the same 
destination. Ha) x. 13° 12' 20" e., 32. 30 miles; 

_(b) jr. 11° 40' 37" B., 32.12 miles. 

8. Find the elements of the great circle track between New 
York light and Cape Clear, 51° 26' n., 9°29'w. 

9. So, between San Francisco, 37°48'n., 122° 25' w., and 
Cape of Good Hope, 33°56's., 18°29'e. 



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